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ELEMENTS 


%«de'r% 


GEOMETRY; 


PHACTICAL     A  V^Jj  I  G:A%i  QjNS 


MENSURATION. 


Jbr  BENJAMIN  GKEENLEAF,  A.  M., 

AOTHOS  OF    "THK    NATIONAL    ARIxmCBTIC,"    "  TREATI8K    ON    ALQTBMJJ*   «XC. 


Nineteenth    Electrotype    Edition.      r*fcrJ^  "^ 


fi?in^ 


BOSTON: 

PUBLISHED  BY  ROBERT   S.  DAVIS  &   CO. 

NEW  YORK:   OAKLEY  AND  MASON,   AND   W.   I.   POOLKY. 
PHILA.DELPHIA :   J.   B.   LIPPINCOTT  AND   COMPANY. 
CINCINNATI:    G.    8.   BLANCHARD   fc   CO- 
ST.  LOUIS:     KEITH   AND  WOODS. 

1868. 


GREENLEAF'S        ^7 
MATHEMATICAL    SERIES. 

I.  Greenleap's  New  Primary  Arithmetic  ;  an  attractive 
book  of  easy  lessons  for  beginners.     84  pp. 

II.  Greenleaf's  New  Intellectual  Arithmetic  ;  a  late 
work,  with  models  of  analysis.     180  pp. 

III.  Greenleaf's  Common  School  Arithmetic  ;  a  com- 
plete «ystem  of*  Writ^eJii  Arithmetic  for  Common  Schools.     324  pp. 

IV.'  (jtREenleaf's  NATiotXAL  ARITHMETIC ;  a  thorough  course 
for-,Hi^  '«ScKools,'Aca3eimee,.* Normal  Schools,  and  Commercial 
Colleges.'    444  pp. 

V.  Greenleaf's  New  Elementary  Algebra  ;  or  the 
First  Principles  of  Analysis  for  Schools  and  Academies.     324  pp. 

VI.  Greenleaf's  New  Higher  Algebra;  an  advanced 
Analytical  Course,,  for  High  Schools  and  Colleges.     394  pp. 

VII.  Greenleaf's  Elements  of  Geometry  ;  with  applica- 
tions to  Mensuration.     320  pp.        L^  ~7  "?■   '■^     ■ 

VIII.  Greenleaf's  Elements  of  Trigonometry;  with 
Practical  Applications  and  Tables.     170  pp. 

IX.  Greenleaf's  Geometry  and  Trigonometry;  or  the 
last  two  works  in  one  volume.     490  pp. 

X.  Greenleaf's  Surveying  and  Navigation,  with  Prac- 
tical Applications  and  Tables.     \In 'pre'paration.\ 

S:^  Keys  to  the  Arithmetics,  Algebras,  Geometry  and 
Trigonometry.     For  Teachers  only.     6  volumes. 


Entered  according  to  Act  of  Congress,  in  tlie  year  1868,  by 

BENJAMIN     GREENLEAF, 

In  the  Clerk's  Office  of  ttie  District  Court  of  tlie  District  of  Massaohosetts. 


KIVEHSIDK,    CAMB'^IDGE: 
PBIMTSP  BT  B.  0.  HOUQHXON  AlfD  OOMFANT. 


PREFACE. 


The  preparation  of  this  treatise  has  been  undertaken  at  the 
earnest  solicitation  of  many  teachers,  who,  having  used  the  au- 
thor's Arithmetics  and  Algebra  with  satisfaction,  have  been  de- 
sirous of  seeing  his  series  rendered  more  complete  by  the  addition 
of  the  Elements  of  Geometry. 

That  there  are  peculiar  advantages  in  a  graded  series  of  text- 
books on  the  same  subject,  few,  if  any,  properly  qualified  to 
judge,  will  doubt.  The  author,  therefore,  feels  justified  in  intro- 
ducing this  volume  to  the  attention  of  the  public. 

In  common  with  most  compilers  of  the  present  day,  he  has 
followed,  in  the  main,  the  simple  and  elegant  order  of  arrange- 
ment adopted  by  Legendre ;  but  in  the  methods  of  demonstra- 
tion no  particular  authority  has  been  closely  followed,  the  aim 
having  been  to  adapt  the  work  fully  to  the  latest  and  most  ap- 
proved modes  of  instruction.  In  this  respect,  it  is  believed,  there 
will  be  found  incorporated  a  considerable  number  of  important 
improvements. 

More  attention  than  is  usual  in  elementary  works  of  this  kind 
has  been  given  to  the  converse  of  propositions.  In  almost  all 
cases  where  it  was  possible,  the  converse  of  a  proposition  has 
been  demonstrated. 

The  demonstration  of  Proposition  XX.  of  tiie  first  book  is 
essentially  the  one  given  by  M.  da  Cunha  in  the  Principes  Mathi- 


IV  PREFACE. 

matiques,  which  has  justly  been  pronounced  by  the  highest 
mathematical  authorities  to  be  a  very  important  improvement  in 
elementary  geometry.  It  has,  however,  never  before  been  intro- 
duced into  a  text-book  by  an  American  author. 

The  Application  of  Geometry  to  Mensuration,  given  in  the 
eleventh  and  twelfth  books,  are  designed  to  show  how  the  theo- 
retical principles  of  the  science  are  connected  with  manifold 
practical  results. 

The  Miscellaneous  Geometrical  Exercises,  which  follow,  are 
calculated  to  test  the  thoroughness  of  the  scholar's  geometrical 
knowledge,  besides  being  especially  adapted  to  develop  skill  and 
discrimination  in  the  demonstration  of  theorems  and  the  solution 
of  problems  unaided  except  by  principles. 

Sufficient  Applications  of  Algebra  to  Geometry  are  given  to 
show  the  relation  existing  between  these  two  branches  of  the 
mathematics.  The  problems  introduced  in  connection  therewith 
will  be  found  to  be,  not  only  of  a  highly  interesting  character, 
but  well  calculated  to  secure  valuable  mental  discipline. 

In  the  preparation  of  this  work  the  author  has  received  valua- 
ble suggestions  from  many  eminent  teachers,  to  whom  he  would 
here  express  his  sincere  thanks.  Especially  would  he  acknowl- 
edge his  great  obligations  to  H.  B.  Maglathlin,  A.  M.,  who  for 
many  months  has  been  associated  with  him  in  his  labors,  and 
to  whose  experience  as  a  teacher,  skill  as  a  mathematician,  and 
ability  as  a  writer,  the  value  of  this  treatise  is  largely  due. 

BENJAMIN  GREENLEAF. 

Bradford,  Mass.,  June  25,  1858. 


NOTICE. 

A  Key,  comprising  the  Solutions  of  the  Problems  contained  in  the 
last  four  Books  of  this  Geometry,  has  been  published,  for  Teachers 
only ;  and  the  same  will  be  mailed,  post-paid,  to  the  address  of  any 
Teacher  who  will  forward  fifty  cents  in  stamps  to  the  Publishers. 


C  ONTENTS 


PLANE     GEOMETRY. 
BOOK    I. 

MOP 
BLEMENTARY  PRINCIPLES  .  .....  7 

BOOK      II. 
RATIO   AND   PROPORTION <9 

BOOK      III. 
THE   CIRCLE,   AND    THE    MEASURE   OF    ANGLES  ...         55 

BOOK    IV. 

PROPORTIONS,   AREAS,   AND   SIMILARITY  OP   FIGURES        .  .         76 

BOOK      V. 
PROBLEMS   RELATING   TO   THE   PRECEDING  BOOKS     .  .  .118 

BOOK    VI. 

BEGULAR   POLYGONS,   AND   THE   AREA   OF   THE   CIRCLE    .  .      148 

1* 


Tl  CONTENTS. 

SOLID     GEOMETRY. 
BOOK    VII. 

PLANES.  —  DIEDRAL   AND   POLYEDRAL    ANGLES  .  .  .      165 

BOOK    VIII. 

POLYEDRONS 184 

BOOK    IX. 

THE   SPHERE,   AND   ITS   PROPERTIES 214 

BOOK      X. 
THE   THREE   ROUND   BODIES .      238 


MENSURATION. 
BOOK    XI. 

j».^PLlCATIONS     OF      GEOMETRY     TO     THE     MENSURATION     OP 

PLANE   FIGURES 253 

BOOK    XII.. 

APPLICATIONS      OF     GEOMETRY     TO     THE     MENSURATION     OF 

SOLIDS '-^8 


BOOK    XIII. 

MISCELLANEOUS   GEOMETRICAL   EXERCISES       .  .  •      801 

BOOK    XIV. 

APPLICATION  OP  ALGEBRA   TO   GEOMETRY  .  .  .811 


ELEMENTS  TTr^^lTOMETRY. 


BOOK   I. 

ELEMENTARY    PRINCIPLES. 
DEFINITIONS. 

1.  Geometry  is  the  science  of  Position  and  Extension. 
The  elements  of  position  are  direction  and  distance. 
The  dimensions  of  extension  are  length,  breadth,  and 

height  or  thickness. 

2.  Magnitude,  in  general,  is  that  which  has  one  or 
more  of  the  three  dimensions  of  extension. 

3.  A  Point  is  that  which  has  position,  without  magni- 
tude. 

4.  A  Line  is  that  which  has  length,  without  either 
breadth  or  thickness. 

5.  A  Straight  Line,  or  Right 

Line,  is  one  which  has  the  same       . 

direction  in  its  whole  extent;  as 
the  line  A  B. 

The  word  line  is  frequently  used  alone,  to  designate  a 
straight  line. 

6.  A  Curved  Line  is  one  which  ^^^     ..^^^ 
continually  changes  its  direction  ;      C  ^^^  ^""^  D 
as  the  line  CD. 

The  word  curve  is  frequently  used  to  designate  a  curved 
line. 


8 


ELEMENTS   OP   GEOMETRY. 


7.  A  Broken  Line  is  one  which  is 
composed  of  straight  lines,  not  lying  in 
the  same  direction  ;  as  the  line  E  F. 

8.  A  Mixed  Line  is  one  which  is  composed  of  straight 
lines' kiid  of  curved  Inies. 

9.  A  Surface  is  that  which  has  length  and  breadth, 
witiiout  height  or  thickness. 

10.  A  Plane  Surface,  or  simply  a  Plane,  is  one  in 
which  any  two  points  being  taken,  the  straight  line  that 
joins  them  will  lie  wholly  in  the  surface. 

11.  A  Curved  Surface  is  one  that  is  not  a  plane  sur- 
face, nor  made  up  of  plane  surfaces. 

12.  A  Solid,  or  Volume,  is  that  which  has  length, 
breadth,  and  thickness. 


ANGLES  AND  LINES. 

13.  A  Plane  Angle,  or  simply  an 
Angle,  is  the  difference  in  the  direc- 
tion of  two  lines,  which  meet  at  a 
point ;  as  the  angle  A. 

The  point  of  meeting.  A,  is  tlie  vertex  of  the  angle,  and 
the  lines  AB,  AC  are  the  sides  of  the  angle. 

An  angle  may  be  designated,  not 
only  by  the  letter  at  its  vertex,  as 
C,  but  by  three  letters,  particularly 
when  two  or  more  angles  have  the 
same  vertex  ;  as  the  angle  A  C  D  or  ^ 

D  C  B,  the  letter  at  the  vertex  always  occupying  the  mid- 
dle place. 

The  quantity  of  an  angle  does  not  depend  upon  the 
length,  but  entirely  upon  the  position,  of  the  sides  ;  for  the 
angle  remains  the  same,  however  the  lines  containing  it 
be  increased  or  diminished. 


BOOK   I. 


9 


D 


14.  Two  straight  lines  are  said  to  B 
be  perpendicular  to  each  other,  when 
their  meeting  forms  equal  adjacent 
angles  ;  thus  the  lines  A  B  and  C  D 
are  perpendicular  to  each  other.              ^    ~ 

Two  adjacent  angles,  as  C  AB  and  BAD,  have  a  com 
mon  vertex,  as  A ;  and  a  common  side,  as  A  B. 

15.  A  Right  Angle  is  one  which 
is  formed  hy  a  straight  line  and  a 
perpendicular  to  it ;  as  the  angle 
CAB. 


B 


D 


16.  An  Acute  Angle  is  one  which 
is  less  than  a  right  angle ;  as  the 
angle  D  E  F. 


An  Obtuse  Angle  is  one  which  is 
greater  than  a  right  angle ;  as  the 
angle  EFG. 


Acute   and  obtuse  angles  have  their  sides  oblique  to 
each  other,  and  are  sometimes  called  oblique  angles. 

17.  Parallel  Lines  are  such  as, 
being  in  the  same  plane,  cannot 
meet,  however  far  either  way  both 
of  them  may  be  produced  ;  as  the 
lines  A B,  CD. 

18.  When  a  straight  line,  as 
E  F,  intersects  two  parallel  lines, 
as  AB,  CD,  the  angles  formed 
by  the  intersecting  or  secant  line 
take  particular  names,  thus  :  — 

Interior  Angles  on  the  same 
Side  are  those  w^liich  lie  within 
the    parallels,   and  on   the  same 


B 


10  ELEMENTS   OP   GEOMETRY. 

side  of  the  secant  line ;  as  the 
angles  BGH,  GHD,  and  also 
AGH,  GHC.  A- 

ExTERiOR  Angles  on  the  same 
Side  are  those  which  lie  without 
the  parallels,  and  on  the  same  side 
of  the  secant  line ;  as  the  angles 
BGE,  DHF,  and  also  the  angles 
AGE,  CHF. 

Alternate  Interior  Angles  lie  within  the  parallels, 
and  on  different  sides  of  the  secant  line,  but  are  not  adja- 
cent to  each  other;  as  the  angles  BGH,  GHC,  and  also 
AGH,   GHD. 

Alternate  Exterior  Angles  lie  without  the  parallels, 
and  on  different  sides  of  the  secant  line,  but  not  adjacent 
to  each  other ;  as  the  angles  E  G B,  CHF,  and  also  the 
angles  AGE,  DHF. 

Opposite  Exterior  and  Interior  Angles  lie  on  the  same 
side  of  the  secant  line,  the  one  without  and  the  other 
within  the  parallels,  but  not  adjacent  to  each  other  ;  as  the 
angles  E  G B,  GHD,  and  also  EGA,  G H C,  are,  respec- 
tively, the  opposite  exterior  and  interior  angles. 

PLANE  FIGURES. 

19.  A  Plane  Figure  is  a  plane  terminated  on  all  sides 
by  straight  lines  or  curves. 

The  boundary  of  any  figure  is  called  its  perimeter. 

20.  When  the  boundary  lines  are 
straight,  the  space  |hey  enclose  is 
called  a  Rectilineal  Figure,  or 
Polygon  ;  as  the  figure  A  B  C  D  E. 

A  B 

21.  A  polygon  of  three  sides  is  called  a  triangle  ;  one 
of  four  sides,  a  quadrilateral  ;  one  of  five,  a  pentagon  ; 
one  of  six,  a  hexagon;  one  of  seven,  a  heptagon;  one 


BOOK   I. 


11 


of  eight,  an  octagon  ;  one  of  nine,  a  nonagon  ;  one  of 
ten,  a  decagon  ;  one  of  eleven,  an  undecagon  ;  one  of 
twelve,  a  dodecagon  ;  and  so  on. 


22.  An  Equilateral  Triangle  is 
one  which  has  its  three  sides  equal ; 
as  the  triangle  ABC. 


An  Isosceles  Triangle  is  one 
which  has  two  of  its  sides  equal ;  as 
the  triangle  D  E  F. 

A  Scalene  Triangle  is  one  which 
has  no  two  of  its  sides  equal ;  as  the 
triangle  G  H  I. 

^3.  A  Right-angled  Triangle  is 
one  which  lias  a  right  angle  ;  as  the 
triangle  J  K  L. 


The  side  opposite  to  the  right  angle  is  called  tlie  hy- 
pothenuse  ;  as  the  side  J  L. 

24.  An  Acute-angled  Triangle  is  one  which  has  three 
acute  angles  ;  as  the  triangles  ABC  and  D  E  F,  Art.  22. 

An  Obtuse-angled  Triangle  is  one  which  has  an  ob- 
tuse angle  ;  as  the  triangle  G  H  I,  Art.  22. 

Acute-angled  and  obtuse-angled  triangles  are  also  called 
oblique-angled  triangles. 

25.  A  Parallelogram  is  a  quadrilateral  which  has  its 
opposite  sides  parallel. 

26.  A  Rectangle  is  any  parallel- 
ogram whose  angles  are  riglit  angles ; 
as  the  parallelogram  A  B  C  D.  ^ 


12 


ELEMENTS    OF   GEOMETRY. 


A  Square  is  a  rectangle  whose 
sides  are  equal ;  as  the  rectangle 
EFGH. 


27.  A  Rhomboid  is  any  parallelo- 
gram whose  angles  are  not  right  an- 
gles ;  as  the  parallelogram  IJKL. 


G 


K 


A  Rhombus  is  a  rhomboid  whose 
sides  are  equal ;  as  the  rhomboid 
MNOP. 


M^ 


28.  A  Trapezoid  is  a  quadrilateral 
which  has  only  two  of  its  sides  par- 
allel ;  as  the  quadrilateral  R  S  T  U.. 


W 


A  Trapezium  is  a  quadrilateral 
which  has  no  two  of  its  sides  paral- 
lel ;  as  the  quadrilateral  V  WXY. 


29.  A  Diagonal  is  a  line  joining 
the  vertices  of  any  two  angles  which 
are  opposite  to  each  other ;  as  the 
lines  E  C  and  E  B  in  the  polygon 
ABODE. 


30.  A  Base  of  a  polygon  is  the  side  on  which  the  poly- 
gon is  supposed  to  stand.  But  in  the  case  of  the  isosceles 
triangle,  it  is  usual  to  consider  that  side  the  base  which  is 
not  equal  to  either  of  the  other  sides. 

31.  An  equilateral  polygon  is  one  which  has  all  its 
sides  equal.     An  equiangular  poly^n  is  one  which  has 


BOOK   I.  13 

all  its  angles  equal.     A  re^lar  polygon  is  one  which  is 
equilateral  and  equiangular. 

32.  Two  polygons  are  mutually  equilateral,  when  all 
the  sides  of  the  one  equal  the  corresponding  sides  of  the 
other,  each  to  each,  and  are  placed  in  the  same  order. 

Two  polygons  are  mutually  equiangular,  when  all  the 
angles  of  the  one  equal  the  corresponding  angles  of  the 
other,  each  to  each,  and  are  placed  in  the  same  order. 

33.  The  corresponding  equal  sides,  or  equal  angles,  of 
polygons  mutually  equilateral,  or  mutually  equiangular, 
are  called  homologous  sides  or  angles. 

AXIOMS. 

34.  An  Axiom  is  a  self-evident  truth ;  such  as,  — 

1.  Things  which  arc  equal  to  the  same  thing,  are  equal 
to  each  other. 

2.  If  equals  be  added  to  equals,  the  sums  will  be  equal. 

3.  If  equals  be  taken  from  equals,  the  remainders  will 
be  equal. 

4.  If  equals  be  added  to  unequals,  the  sums  will  be 
unequal. 

5.  If  equals  be  taken  from  unequals,  the  remainders 
will  be  unequal. 

6.  Things  which  are  double  of  the  same  thing,  or  of 
equal  things,  are  equal  to  each  other. 

7.  Things  which  are  halves  of  the  same  thing,   or  of 
equal  things,  are  equal  to  each  other. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

10.  A  straight  line  is  the  shortest  line  that  can  be 
drawn  from  one  point  to  another. 

11.  From  one  point  to  another  only  one  straight  line 
can  be  drawn. 

12.  Through  the  same  point  only  one  parallel  to  a 
straight  line  can  be  drawn.  • 

2 


14  ELEMENTS   OF   GEOMETRY. 

13.  All  right  angles  are  equal  to  one  another. 

14.  Magnitudes  which  coincide  throughout  their  whole 
extent,  are  equal. 

POSTULATES. 

35.  A  Postulate  is  a  self-evident  problem  ;  such  as,  — 

1.  That  a  straight  line  may  be  drawn  from  one  point  to 
another. 

2.  That  a  straight  line  may  be  produced  to  any  length. 

3.  That  a  straight  line  may  be  drawn  through  a  given 
point  parallel  to  another  straight  line. 

4.  That  a  perpendicular  to  a  given  straight  line  may  be 
drawn  from  a  point  either  within  or  without  the  line. 

5.  That  an  angle  may  be  described  equal  to  any  given 
angle. 

PROPOSITIONS. 

36.  A  Demonstration  is  a  course  of  reasoning  by  which 
a  truth  becomes  evident. 

37.  A  Proposition  is  something  proposed  to  be  demon- 
strated, or  to  be  performed. 

A  proposition  is  said  to  be  the  converse  of  another, 
when  the  conclusion  of  the  first  is  used  as  the  supposition 
in  the  second. 

38.  A  Theorem  is  something  to  be  demonstrated. 

39.  A  Problem  is  something  to  be  performed. 

40.  A  Lemma  is  a  proposition  preparatory  to  the  dem- 
onstration or  solution  of  a  succeeding  proposition. 

41.  A  Corollary  is  an  obvious  consequence  deduced 
from  one  or  more  propositions. 

42.  A  Scholium  is  a  remark  made  upon  one  or  more 
preceding  propositions. 

43.  An  Hypothesis  is  a  supposition,  made  either  in  the 


BOOK  I. 


15 


enunciation  of  a  proposition,  or  in  the  course  of  a  demon- 
stration. 


Proposition  I. — Theorem. 

44.  The  adjacent  angles  which  one  straight  line  makes 
by  meeting  another  straight  line,  are  together  equal  to 
two  right  angles. 

Let  the  straight  line  D  C  meet                       E 
AB,  making  the  adjacent  angles 
A  C  D,   D  C  B  ;    these   angles   to- 
gether will  be  equal  to  two  right 
angles.  A  . [/ g 

From  the  point  C  suppose  C  E  ^ 

to  be  drawn  perpendicular  to  A  B  ;  then  the  angles  ACE 
and  E  C  B  will  each  be  a  right  angle  (Art.  15) .  But  the 
angle  A  C  D  is  composed  of  the  right  angle  ACE  and  the 
angle  ECD  (Art.  34,  Ax.  9),  and  the  angles  ECD  and 
D  C  B  compose  the  other  right  angle,  E  C  B  ;  hence  the 
angles  AC  D,  D  C  B  together  equal  two  right  angles. 

45.  Cor.  1.  If  one  of  the  angles  A  C  D,  D  C  B  is  a 
right  angle,  the  other  must  also  be  a  right  angle. 

46.  Cor.  2.  All  the  successive 
angles,  B  A  C,  CAD,  DAE, 
EAF,  formed  on  the  same  side 
of  a  straight  line,  B  F,  are  equal, 
when  taken  together,  to  two  right 
angles  ;  for  their  sum  is  equal  to  g 
that  of  the  two  adjacent  angles,  -^ 
BAC,  CAF. 

Proposition  II.  —  Theorem. 

47.  If  one  straight  line  meets  two  other  straight  lines 
at  a  common  point,  making  adjacent  angles,  ivhich  to- 
gether are  equal  to  two  right  angles,  the  two  lines  form 
one  and  the  same  straight  line. 


16 


ELEMENTS   OF   GEOMETRY. 


Let  the  straight  line  D  C  meet 
the  two  straight  lines  AC,  C  B  at 
the  common  point  C,  making  the 
adjacent  angles  A  C  D,  D  C  B  to- 
gether equal  to  two  right  angles  ;  

then  the  lines  AC   and  CB  will      ^  C     z:;::;;      ^ 

- -p 

form  one  and  the  same  straight 
line. 

If  C  B  is  not  the  straight  line  A  C  produced,  let  C  E  be 
that  line  produced  ;  then  the  line  ACE  being  straight, 
the  sum  of  the  angles  A  C  D  and  D  C  E  will  be  equal  to 
two  right  angles  (Prop.  I.).  But  by  hypothesis  the  angles 
A  C  D  and  D  C  B  are  together  equal  to  two  right  angles  ; 
therefore  the  sum  of  the  angles  A  C  D  and  D  C  E  must  be 
equal  to  the  sum  of  the  angles  A  C  D  and  D  C  B  (Art.  34, 
Ax.  2).  Take  away  the  common  angle  A  C  D  from  each, 
and  there  will  remain  the  angle  D  C  B,  equal  to  the  angle 
D  C  E,  a  part  to  the  whole,  which  is  impossible  ;  therefore 
C  E  is  not  the  line  A  C  produced.  Hence  A  C  and  C  B 
form  one  and  the  same  straight  line. 


Proposition  III. — Theorem, 


48.  Two  straight  lines,  ivhich  have  tivo  points  co^nmon, 
coincide  ivith  each  other  throughout  their  ivhole  extent^ 
and  form  one  and  the  same  straight  line. 

Let  the  two  points  which  are  F 

common  to  two  straight  lines  be 
A  and  B. 

The   two    lines   must   coincide 

between  tlie  points  A  and  B,  for     a 

otherwise    there   would    be    two  ^    ^ 

straight  lines  between  A  and  B,  which  is  impossible  (Art. 

34,  Ax.  11). 

Suppose,  however,  that,  on  being  produced,  the  lines 
begin  to  separate  at  the  point  C,  the  one  taking  the  direc- 


BOOK   I;  17 

tion  C  D,  and  the  other  C  E.  From  the  point  C  let  the 
line  C  F  be  drawn,  making,  with  C  A,  the  right  angle 
A  C  F.  Now,  since  A  C  D  is  a  straight  line,  the  angle 
F  C  D  will  be  a  right  angle  (Prop.  I.  Cor.  1)  ;  and  since 
ACE  is  a  straight  line,  the  angle  F  C  E  will  also  be  a 
right  angle ;  therefore  the  angle  F  C  E  is  equal  to  the 
angle  F  C  D  (Art.  34,  Ax.  13),  a  part  to  the  whole,  which 
is  impossible  ;  hence  two  straight  lines  which  have  two 
points  common,  A  and  B,  cannot  separate  from  each  other 
when  produced ;  hence  they  must  form  one  and  the  same 
straight  line. 

Proposition  TV.  —  Theorem. 

49.  When  ttvo  straight  lines  intersect  each  other,  the 
opposite  or  vertical  angles  which  they  form  are  equal. 

Let  the  two  straight  lines  A  B,  ^ 

C  D  intersect  each  other  at  the 
point  E ;  then  will  the  angle  AE  C 
be  equal  to  the  angle  DEB,  and 
the  angle  CEB  to  AED. 

For  the  angles  AEC,   CEB,  D 

which  the  straight  line  C  E  forms  by  meeting  the  straight 
line  AB,  are  together  equal  to  two  right  angles  (Prop.  I.)  ; 
and  the  angles  CEB,  BED,  which  the  straight  line  B E 
forms  by  meeting  the  straight  line  C  D,  are  equal  to  two 
right  angles  ;  hence  the  sum  of  the  angles  AEC,  CEB 
is  equal  to  the  sum  of  the  angles  CEB,  BED  (Art.  34, 
Ax.  1).  Take  away  from  each  of  these  sums  the  common 
angle  CEB,  and  there  will  remain  the  angle  AEC,  equal 
to  its  opposite  angle,  BED  (Art.  34,  Ax.  3). 

In  the  same  manner  it  may  be  shown  that  the  angle 
C E B  is  equal  to  its  opposite  angle,  AED. 

50.  Cor.  1.  The  four  angles  formed  by  two  straight 
lines  intersecting  each  other,  are  together  equal  to  four 
right  angles. 

2* 


18  ELEMENTS   OF   GEOMETRY. 

51.  Cor.  2.  All  the  successive  angles,  around  a  com- 
mon point,  formed  by  any  number  of  straight  lines  meet- 
ing at  that  point,  are  together  equal  to  four  right  angles. 

Proposition  Y.  —  Theorem. 

52.  If  two  triangles  have  tioo  sides  and  the  included 
angle  in  the  one  equal  to  two  sides  and  the  included  angle 
in  the  other,  each  to  each,  the  two  triangles  will  be  equal. 

In  the  two  triangles 
ABC,  DEF,  let  the 
side  A  B  be  equal  to  the 
side  DE,  the  side  AC 
to  the  side  D  F,  and  the 
angle  A  to  the  angle  D  ; 
then  the  triangles  ABC,  DEF  will  be  equal.      ^ 

Conceive  the  triangle  ABC  to  be  applied  to  the  triangle 
D  EF,  so  that  the  side  AB  shall  fall  upon  its  equal,  D  E, 
the  point  A  upon  D,  and  the  point  B  upon  E  ;  then,  since 
the  angle  A  is  equal  to  the  angle  D,  the  side  AC  will  take 
the  direction  DF.  But  AC  is  equal  to  D  F  ;  therefore  the 
point  C  will  fall  upon  F,  and  the  third  side  B  C  will  co- 
incide with  the  third  side  E  F  (Art.  34,  Ax.  11).  Hence 
tlie  triangle  ABC  coincides  with  the  triangle  DEF,  and 
they  are  therefore  equal  (Art.  34,  Ax.  14). 

53.  Cor.  When,  in  two  triangles,  these  three  parts  are 
equal,  namely,  the  side  A  B  equal  to  D  E,  the  side  A  C 
equal  to  D  F,  and  the  angle  A  equal  to  D,  the  other  three 
corresponding  parts  are  also  equal,  namely,  the  side  B  C 
equal  to  E  F,  the  angle  B  equal  to  E,  and  the  angle  C 
equal  to  F. 

Proposition  YI.  —  Theorem. 

54.  If  two  triangles  have  two  angles  and  the  included 
side  in  the  one  equal  to  two  angles  and  the  included  side 
in  the  other,  each  to  each,  the  two  triangles  will  he  equal. 


BOOK   L 


19 


In  the  two  triangles 
ABC,  DEF,  let  the 
angle  B  be  equal  to  the 
angle  E,  the  angle  C  to 
the  angle  F,  and  the  side 
BC  to  the  side  EF; 
then  the  triangles  ABC,  DEF  will  be  equal. 

Conceive  the  triangle  A  B  C  to  be  applied  to  the  triangle 
DEF,  so  that  the  side  B  C  shall  fall  upon  its  equal,  E F, 
the  point  B  upon  E,  and  the  point  C  upon  F.  Then, 
since  the  angle  B  is  equal  to  the  angle  E,  the  side  B  A 
will  take  the  direction  E  D  ;  therefore  the  point  A  will  be 
found  somewhere  in  the  line  ED.  In  like  manner,  since 
the  angle  C  is  equal  to  the  angle  F,  the  line  C  A  will  take 
the  direction  F  D,  and  the  point  A  will  be  found  some- 
where in  the  Ime  F  D.  Hence  the  point  A,  falling  at  the 
same  time  in  both  of  the  straight  lines  E  D  and  F  D,  must 
fall  at  their  intersection,  D.  Hence  the  two  triangles 
ABC,  DEF  coincide  with  each  other,  and  are  therefore 
equal  (Art.  34,  Ax.  14). 

55.  Cor,  When,  in  two  triangles,  these  three  parts  are 
equal,  namely,  the  angle  B  equal  to  the  angle  E,  the  angle 
C  equal  to  tlie  angle  F,  and  the  side  B  C  equal  to  the  side 
EF,  the  other  three  corresponding  parts  are  also  equal; 
namely,  the  side  B  A  equal  to  E  D,  the  side  C  A  equal  to 
F  D,  and  the  angle  A  equal  to  the  angle  D. 

Proposition  YII,  —  Theorem. 

56.  In  an  isosceles  triang'le,  the  angles  opposite  the 
equal  sides  are  equal. 

Let  ABC  be  an  isosceles  triangle,  in 
which  the  side  AB  is  equal  to  the  side 
A  C ;  then  will  the  angle  B  be  equal  to 
the  angle  C. 

Conceive  the  angle  B  A  C  to  be  bisect- 
ed, or  divided  into  two  equal  parts,  by     ^ 


20 


ELEMENTS  OF    GEOMETRY. 


the  straight  line  AD,  making  the  angle 

BAD  equal  to  D  A  C.     Then  the  two 

triangles   BAD,    CAD    have   the   two 

sides  A  B,  A  D   and  the  included  angle 

in  the  one  equal  to  the  two  sides  A  C, 

A  D  and  the  included  angle  in  the  other,     B 

each  to  each  ;  hence  the  two  triangles  are  equal,  and  the 

angle  B  is  equal  to  the  angle  C  (Prop.  V.). 

57.  Cor.  1.  The  line  bisecting  the  vertical  angle  of  an 
isosceles  triangle  bisects  the  base  at  right  angles. 

58.  Cor.  2.  Conversely,  the  line  bisecting  the  base  of 
an  isosceles  triangle  at  right  angles,  bisects  also  the  verti- 
cal angle. 

59.  Cor.  3.  Every  equilateral  triangle  is  also  equian- 
gular. 

Proposition  YIII.  —  Theorem. 

60.  ff  two  angles  of  a  triangle  are  equal,  the  opposite 
sides  are  also  equal,  and  the  triangle  is  isosceles. 

Let  A  B  C  be  a  triangle  having  the  an-  A 

gle  B  equal  to  the  angle  C ;  then  will  the 
side  A  B  be  equal  to  the  side  A  C. 

For,  if  the  two  sides  are  not  equal,  one 
of  them  must  be  greater  than  the  other. 
Let  A  B  be  the  greater ;  then  take  D  B 
equal  to  AC  the  less,  and  draw  CD.    -^  ^ 

Now,  in  the  two  triangles  D  B  C,  A  B  C,  we  have  D  B 
equal  to  A  C  by  construction,  the  side-  B  C  common,  and 
the  angle  B  equal  to  the  angle  A  C  B  by  hypothesis  ; 
therefore,  since  two  sides  and  the  included  angle  in  the 
one  are  equal  to  two  sides  and  the  included  angle  in  the 
other,  each  to  each,  the  triangle  D  B  C  is  equal  to  the 
triangle  ABC  (Prop.  Y.),  a  part  to  the  whole,  wliich  is 
impossible  (Art.  34,  Ax.  8).  Hence  tlie  sides  AB  and 
A  C  cannot  be  unequal ;  therefore  the  triangle  A  B  C  is 
isosceles. 


BOOK   I. 


21 


61.    Cor. 
lateral. 


Therefore  every  equiangular  triangle  is  equi- 


B 


Proposition  IX.  —  Theorem. 

62.  Any  side  of  a  triangle  is  less  than  the  sum  of  the 
other  two. 

In  the  triangle  ABC,  any  one  side,  C 

as  AB,  is  less  than  the  sum  of  the  other 
two  sides,  A  C  and  C  B. 

For   the   straight    line    AB    is   the 
shortest  line  that  can  be  drawn  from 
the  point  A  to  the  point  B  (Art.  34, 
Ax.  10)  ;  hence  the  side  A  B  is  less  than  the  sum  of  the 
sides  A  C  and  C  B. 

In  like  manner  it  may  be  proved  that  the  side  A  C  is 
less  than  the  sum  of  A  B  and  B  C,  and  the  side  B  C  less 
than  the  sum  of  B  A  and  A  C. 

63.  Cor.  Since  the  side  AB  is  less  than  the  sum  of  AC 
and  C  B,  if  we  take  away  from  each  of  tliese  two  unequal s 
the  side  C  B,  we  shall  have  the  difference  between  A  B 
and  C  B  less  than  A  C  ;  that  is,  the  difference  between 
any  tivo  sides  of  a  triangle  is  less  than  the  other  side. 

Proposition  X.  —  Theorem. 

64.  The  greater  side  of  any  triangle  is  opposite  the 
greater  angle. 

In  the  triangle  CAB,  let  the  angle      A 
C  be  greater  than  B  ;  then  will  the  side 
A  B,  opposite   to  C,   be    greater   than 
A  C,  opposite  to  B. 

Draw  the  straight  line  C  D,  making 
the  angle  BCD  equal  to  B.     Then,  in 
the  triangle  B  D  C,  we  shall  have  the 
side  BD  equal  to  D  C  (Prop.  VIII.).     But  the  side  A  C 
is  less  than  the  sum  of  AD  and  D  C  (Prop.  IX.),  and  the 


22  ELEJVIENTS   OF   GEOMETRY. 

sum  of  A  D  and  D  C  is  equal  to  the  sum  of  A  D  and  D  B, 
which  is  equal  to  A  B  ;  therefore  the  side  A  B  is  greater 
than  AC. 

65.  Cor.  1.  Therefore  the  shorter  side  is  opposite  to  the 
less  angle. 

66.  Cor.  2.  In  the  right-angled  triangle  the  hypothe- 
nuse  is  the  longest  side. 


Proposition  XI.  —  Theorem. 

67.  The  greater  angle  of  any  triangle  is  opposite  the 
greater  side. 

In  the  triangle  CAB,  suppose  the      a 
side  A  B  to  be  greater  than  A  C  ;  then 
will  the  angle  C,  opposite  to  A  B,  be 
greater  than  the  angle  B,  opposite  to 
AC. 

For,  if  the  angle  C  is  not  greater  than 
B,  it  miist  either  be  equal  to  it  or  less. 
If  the  angle  C  were  equal  to  B,  then  would  the  side  AB 
be  equal  to  the  side  AC  (Prop.  VIII.),  which  is  contrary 
to  the  hypothesis  ;  and  if  the  angle  C  were  less  than  B, 
then  would  the  side  AB  be  less  than  AC  (Prop.  X. 
Cor.  1),  which  is  also  contrary  to  the  hypothesis.  Hence, 
the  angle  C  must  be  greater  than  B. 

68.  Cor.  It  follows,  therefore,  that  the  less  angle  is 
opposite  to  the  shorter  side. 

Proposition  XII.  —  Theorem, 

69.  If,  from  any  point  within  a  triangle,  tioo  straight 
lines  are  draivn  to  the  extremities  of  either  side,  their 
sum  will  he  less  than  that  of  the  other  two  sides  of  the 
triangle. 


BOOK  I. 


23 


Let  the  two  straight  lines  B  0,  C  0 
be  drawn  from  the  point  O,  within  the 
triangle  ABC,  to  the  extremities  of 
the  side  B  C  ;  then  will  the  sum  of  the 
two  lines  B  0  and  0  C  be  less  than 
the  sum  of  the  sides  B  A  and  A  C. 

Let  the  straight  line  BO  be  pro- 
duced till  it  meets  the  side  A  C  in  the  point  D  ;  and  be- 
cause one  side  of  a  triangle  is  less  than  the  sum  of  the 
other  two  sides  (Prop.  IX.),  the  side  OC  in  the  triangle 
C  D  0  is  less  than  the  sum  of  0  D  and  DC.  To  each  of 
these  inequalities  add  B  0,  and  we  have  the  sum  of  B  O 
and  0  C  less  than  the  sum  of  BO,  OD,  and  DC  (Art.  34, 
Ax.  4)  ;  or  the  sum  of  B  0  and  0  C  less  than  the  sum  of 
B  D  and  D  C.  Again,  because  the  side  B  D  is  less  than 
the  sum  of  B  A  and  A  D,  by  adding  D  C  to  each,  we  have 
the  sum  of  B  D  and  D  C  less  than  the  sum  of  B  A  and 
A  C.  But  it  has  been  just  shown  that  the  sum  of  B  0  and 
0  C  is  less  than  the  sum  of  B  D  and  D  C  ;  much  more, 
then,  is  the  sum  of  B  0  and  0  C  less  than  B  A  and  A  C. 


Proposition  XIII.  —  Theorem. 

70.  From  a  point  ivithout  a  straight  line,  only  one  per- 
pendicular can  be  drawn  to  that  line. 

Let  A  be  the  point,  and  DE  the 
given  straight  line  ;  then  from  the 
point  A  only  one  perpendicular  can 
be  drawn  to  D  E: 

Let  it  be  supposed  that  we  can 
draw  two  perpendiculars,  AB  and 
AC.  Produce  one  of  them,  as  AB, 
till  BF  is  equal  to  AB,  and  join  F  C. 
Then,  in  the  triangles  ABC  and  C  B  F,  the  angles  C  B  A 
and  CBF  are  both  right  angles  (Prop.  I.  Cor.  1),  the 
side  C  B  is  common  to  both,  and  the  side  B  F  is  equal  to 


24 


ELEMENTS    OF   GEOMETRY. 


the  side  A  B  ;  hence  the  two  triangles 
are  equal,  and  the  angle  B  C  F  is  equal 
to  the  angle  B  C  A  (Prop.  Y .)  But 
the  angle  BCA  is,  by  hypothesis,  a 
right  angle ;  therefore  B  C  F  must  also 
be  a  right  angle  ;  and  if  the  two  adja- 
cent angles,  BCA  and  B  C  F,  are  to- 
gether equal  to  two  right  angles,  the 
two  lines  A  C  and  C  F  must  form  one  and  the  same 
straight  line  (Prop.  II.).  Whence  it  follows,  that  be- 
tween the  same  two  points,  A  and  F,  two  straight  lines 
can  be  drawn,  which  is  impossible  (Art.  34,  Ax.  11)  ; 
hence  no  more  than  one  perpendicular  can  be  drawn  from 
the  same  point  to  the  same  straight  line. 

71.  Cor.  At  the  same  point  C,  in  the 
line  AB,  it  is  likewise  impossible  to 
erect  more  than  one  perpendicular  to 
that  line.  For,  if  C  D  and  C  E  were 
each  perpendicular  to  A  B,  the  angles 
BCD,  B  C E  would  be  right  angles  '/^  ^ 

hence  the  angle  BCD  would  be  equal  to  the  angle  B  C  E, 
a  part  to  the  whole,  which  is  impossible. 


E 


D 

/ 


/ 


Proposition  XIY.  —  Theorem. 


72.  If^  from  a  point  without  a  straight  line,  a  perpen- 
dicular be  let  fall  on  that  line,  and  oblique  lines  be  drawn 
to  different  points  in  the  same  line  ;  — 

1st.  The  perpendicular  will  be  shorter  than  any  oblique 
line. 

2d.  Any  tiuo  oblique  lines,  which  meet  the  given  line 
at  equal  distances  from  the  perpendicular,  will  be  equal. 

3d.  Of  any  two  oblique  lines,  that  vjhich  meets  the 
given  line  at  the  greater  distance  from  the  perpendicular 
will  be  the  longer. 


BOOK   I. 


25 


Let  A  be  the  given  point,  an4. 
D  E  the  given  straight  line. 
Draw  A  B  perpendicular  to  D  E, 
and  the  oblique  lines  AE,  AC, 
AD.  Produce  AB  till  BF  is 
equal  to  A  B,  and  join  C  F,  D  F. 

First.   The  triangle  B  C  F  is 
equal  to  the  triangle  B  C  A,  for  -^ 

they  have  the  side  C  B  common,  the  side  A  B  equal  to  the 
side  B  F,  and  the  angle  ABC  equal  to  the  angle  F  B  C, 
both  being  right  angles  (Prop.  I.  Cor.  1)  ;  hence  the  third 
sides,  CF  and  AC,  are  equal  (Prop.  V.  Cor.).  But 
A  B  F,  being  a  straight  line,  is  shorter  than  A  C  F,  which 
is  a  broken  line  (Art.  34,  Ax.  10)  ;  therefore  A  B,  the 
half  of  A  B  F,  is  shorter  than  A  C,  the  half  of  A  C  F  ; 
hence  the  perpendicular  is  shorter  than  any  oblique  line. 

Secondly.  If  B  E  is  equal  to  B  C,  then,  since  A  B  is 
common  to  the  triangles,  ABE,  ABC,  and  the  angles 
ABE,  ABC  are  right  angles,  the  two  triangles  are  equal 
(Prop.  Y.),  and  the  side  A  E  is  equal  to  the  side  AC 
(Prop.  V.  Cor.).  Hence  the  two  oblique  lines,  meeting 
the  given  line  at  equal  distances  from  the  perpendicular, 
are  equal. 

Thirdly.  The  point  C  being  in  the  triangle  ADF,  the 
sum  of  the  lines  A  C,  C  F  is  less  than  the  sum  of  the  sides 
AD,  DF  (Prop.  XII.)  But  AC  has  been  shown  to  be 
equal  to  C  F  ;  and  in  like  manner  it  may  be  shown  that 
A  D  is  equal  to  D  F.  Therefore  A  C,  the  half  of  the  line 
A  C  F,  is  shorter  than  AD,  the  half  of  the  line  ADF; 
hence  the  oblique  line  which  meets  the  given  line  the 
greater  distance  from  the  perpendicular,  is  the  longer. 

73.  Cor.  1,  The  perpendicular  measures  the  shortest 
distance  of  any  point  from  a  straight  line. 

74.  Cor.  2.  From  the  same  point  to  a  giren  straight 
line  only  two  equal  straight  lines  can  be  drawn. 


26  ELEMENTS   OF   GEOMETRY. 

75.  Cor.  3.  Of  any  two  straight  lines  drawn  from  a 
point  to  a  straight  line,  that  whicli  is  not  shorter  than  the 
other  will  be  longer  than  any  straight  line  that  can  be 
drawn  between  them,  from  the  same  point  to  the  same 
line. 

Proposition  XY.  —  Theorem. 

76.  If  from  the  middle  point  of  a  straight  line  a  per- 
pendicular to  this  line  be  drawn, — 

1st.  Any  point  in  the  perpendicular  will  he  equally  dis- 
tant from  the  extremities  of  the  line. 

2d.  Any  point  out  of  the  perpendicular  will  be  un- 
equally distant  from  those  extremities. 

Let  D  C  be  drawn  perpendicular  to 
the  straight  line  A  B,  from  its  middle 
point  C. 

First.  Let  D  and  E  be  points,  taken 
at  pleasure,  in  the  perpendicular,  and 
join  DA,  DB,  and  also  AE,  EB. 
Then,  since  A  C  is  equal  to  C  B,  the 
two  oblique  lines  D  A,  D  B  meet  points  which  are  at  the 
same  distance  from  the  perpendicular,  and  are  therefore 
equal  (Prop.  XIY.).  So,  likewise,  the  two  oblique  lines 
E  A,  E  B  are  equal ;  therefore  any  point  in  the  perpendic- 
ular is  equally  distant  from  the  extremities  A  and  B. 

Secondly.  Let  F  be  any  point  out  of  the  perpendicular, 
and  join  F  A,  F  B.  Then  one  of  those  lines  must  cut  the 
perpendicular,  in  some  point,  as  E.  Join  E  B  ;  then  we 
have  EB  equal  to  EA.  But  in  the  triangle  F  E  B,  the 
side  F  B  is  less  than  the  sum  of  the  sides  E  F,  E  B  (Prop. 
IX.),  and  since  the  sum  of  FE,  E  B  is  equal  to  the  sum 
of  F  E,  E  A,  which  is  equal  to  F  A,  F  B  is  less  than  F  A. 
Hence  any  point  out  of  the  perpendicular  is  at  unequal 
distances  from  the  extremities  A  and  B. 

77.  Cor.  If  a  straight  line  have  two  points,  of  whicli 
each  is  equally  distant  from  the  extremities  of  another 


BOOIv    I. 


27 


straight  line,  it  will  be  perpendicular  to  that  line  at  its 
middle  pohit. 


Proposition  XYI.  —  Theorem. 

78.  If  tivo  triangles  have  tioo  sides  of  the  one  equal  to 
two  sides  of  the  other,  each  to  each,  and  the  included  an- 
gle of  the  one  greater  than  the  included  angle  of  the  other, 
the  third  side  of  that  which  has  the  greater  angle  will  be 
greater  than  the  third  side  of  the  other. 

Let  ABC,  DEF 
be  two  triangles, 
having  the  side 
A  B  equal  to  D  E, 
and  AC  equal  to 
D  F,  and  the  angle 
A  greater  than  D  ;  ^ 
then  will  the  side 
BC  be  greater  than  EF. 

Of  the  two  sides  D  E,  D  F,  let  D  F  be  the  side  which  is 
not  shorter  than  the  other  ;  make  the  angle  E  D  G  equal 
to  B  A  C  ;  and  make  D  G  equal  to  A  C  or  D  F,  and  join 
EG,  GF. 

Since  D  F,  or  its  equal  D  G,  is  not  shorter  than  D  E,  it 
is  longer  than  D  H  (Prop.  XIY.  Cor.  3)  ;  therefore  its 
extremity,  F,  must  fall  below  the  line  E  G.  The  two  tri- 
angles, ABC  and  D  E  G,  have  the  two  sides  AB,  AC 
equal  to  the  two  sides  D  E,  D  G,  each  to  each,  and  the 
'.eluded  angle  BAC  of  the  one  equal  to  the  included 
angle  E  D  G  of  the  other ;  hence  the  side  B  C  is  equal 
to  EG  (Prop.  Y.  Cor.). 

In  the  triangle  D  F  G,  since  D  G  is  equal  to  D  F,  the 
angle  D  F  G  is  equal  to  the  angle  D  G  F  (Prop.  YII.)  ; 
but  the  angle  D  G  F  is  greater  than  the  angle  E  G  F  ; 
therefore  the  angle  D  F  G  is  greater  than  E  G  F,  and 
much  more  is  the  angle  EFG  greater  than  the  angle 


28  ELEMENTS    OF   GEOMETRY. 

E  G  F.  Because  tlie  angle  E  F  G  in  the  triangle  E  F  G 
is  greater  than  EGF,  and  because  the  greater  side  is 
opposite  the  greater  angle  (Prop.  X.),  the  side  E  G  is 
greater  than  E  F ;  and  E  G  has  been  shown  to  be  equal  to 
B  C  ;  hence  B  C  is  greater  than  E  F. 

Proposition  XYII.  —  Theorem.     . 

79.  If  two  triangles  have  tivo  sides  of  the  one  equal  to 
tiuo  sides  of  the  other ^  each  to  each,  but  the  third  side  of 
the  one  greater  than  the  third  side  of  the  other,  the  angle 
coyitained  by  the  sides  of  that  ivhich  has  the  greater  third 
side  ivill  be  greater  than  the  angle  contained  by  the  sides 
of  the  other. 

Let  ABC,  DEF  be 
two  triangles,  the  side 
AB  equal  to  D  E,  and 
AC  equal  to  DF,  and 
the  side  C  B  greater  than 
EF,  then  will  the  angle     b~~  ^q 

A  be  greater  than  D.  ^"^  F 

For,  if  it  be  not  greater,  it  must  either  be  equal  to 
it  or  less.  But  the  angle  A  cannot  be  equal  to  D,  for 
tlicn  the  side  B  C  would  be  equal  to  E  F  (Prop.  Y.  Cor.), 
which  is  contrary  to  the  hypothesis ;  neither  can  it  be 
loss,  for  then  the  side  B  C  would  be  less  than  E  F  (Prop. 
XVI.),  which  also  is  contrary  to  the  hypothesis  ;  there- 
fore the  angle  A  is  not  less  than  the  angle  D,  and  it 
has  been  shown  that  is  not  equal  to  it ;  hence  the  angle 
A  must  be  greater  than  the  angle  D. 

Proposition  XVIII.  —  Theorem. 

80.  If  tioo  triangles  have  the  three  sides  of  the  one 
equal  to  the  three  sides  of  tlie  other,  each  to  each,  the 
triangles  themselves  will  be  equal. 


BOOK 


29 


Let  the  triangles  ABC, 
D  E  F  have  the  side  A  B 
equal  to  D  E,  A  C  to  D  F, 
and  B  C  to  E  F  ;  then  will 
the  angle  A  be  equal  to  D, 
the  angle  B  to  the  angle 

E,  and  the  angle  C  to  the  angle  F,  and  tlie  two  triangles 
will  also  be  equal. 

For,  if  the  angle  A  were  greater  than  the  angle  D,  since 
the  sides  A  B,  A  C  are  equal  to  the  sides  D  E,  D  F,  each 
to  each,  the  side  B  C  would  be  greater  than  E  F  (Prop. 
XVI.)  ;  and  if  the  angle  A  were  less  than  D,  it  would 
follow  that  the  side  B  C  would  bo  less  than  E  F.  But  by 
hypothesis  B  C  is  equal  to  E  F ;  hence  tlie  angle  A  can 
neither  be  greater  nor  less  than  D  ;  therefore  it  must  bo 
equal  to  it.  In  the  same  manner,  it  may  be  shown  that 
the  angle  B  is  equal  to  E,  and  the  angle  C  to  F ;  hence 
tlie  two  triangles  must  be  equal. 

81.  Scholium.  In  two  triangles  equal  to  each  other,  the 
equal  angles  are  opposite  the  equal  sides ;  thus  the  equal 
angles  A  and  D  are  opposite  the  equal  sides  B  C  and  EP. 


Proposition  XIX.  —  Theorem. 

82.  If  Uvo  right-angled  triangles  have  the  hypothenuse 
and  a  side  of  the  one  equal  to  the  hypothenuse  and  a  side 
of  the  other,  each  to  each,  the  triangles  are  equal. 

Let    the    two    right-an-     A  D 

gled  triangles  ABC,  DEF, 
have  the  hypothenuse  A  C 
equal  to  D  F,  and  the  side 
A  B  equal  to  D  E  ;  then 
will  the  triangle  A  B  C  be 
equal  to  the  triangle  DEF.     B  G        C     E  F 

The  two  triangles  are  evidently  equal,  if  the  sides  B  C 
and  EF  are  equal  (Prop.  XVIII.) .     If  it  be  possible,  let 

3* 


30 


ELEMENTS   OF   GEOMETRY. 


these  sides  be  unequal,  and 
let  B  C  be  the  greater. 
Take  B  G  equal  to  E  F,  the 
less  side,  and  join  A  G. 
Then,  in  the  two  triangles 
A  B  G,  D  E  F,  the  angles 
B  and  E  are  equal,  both    B  G        C     E  F 

being  right  angles,  the  side  A  B  is  equal  to  D  E  by  hy- 
pothesis, and  the  side  B  G  to  E  F  by  construction  ;  hence 
these  triangles  are  equal  (Prop.  V.)  ;  and  therefore  A  G 
is  equal  to  D  F.  But  by  hypothesis  D  F  is  equal  to  A  C, 
and  therefore  A  G  is  equal  to  A  C.  But  the  oblique  line 
A  C  cannot  be  equal  to  A  G,  which  meets  the  same  straight 
line  nearer  the  perpendicular  A  B  (Prop.  XIV.)  ;  there- 
fore B  C  and  E  F  cannot  be  unequal,  hence  they  must  be 
equal;  therefore  the  triangles  ABC  and  DEF  are  equal. 


E 


Proposition  XX.  —  Theorem. 

SS.  If  a  straight  line,  intersecting  tivo  other  straight 
lines,  makes  the  alternate  angles  equal,  the  two  lines  are 
parallel. 

Let  the  straight  line 
EF  intersect  the  two 
straight  lines  A  B,  CD, 
making  the  alternate  an- 
gles BGH,  CHG  equal;  I 
then  the  lines  A  B,  C  D 
will  be  parallel. 

For,  if  the  lines  A  B, 
C  D  are  not  parallel,  let 
them  meet  in  some  point  K,  and  through  0,  the  middle 
point  of  GH,  draw  the  straight  line  IK,  making  10  equal 
to  0  K,  and  join  H  I.  Then  the  opposite  angles  K  0  G, 
I  0  H,  formed  by  the  intersection  of  the  two  straight  lines 
IK,  GH,  are  equal  (Prop.  IV.)  ;  and  the  triangles  K  0  G, 


K 


BOOK   I.  31 

1 0  H  have  the  two  sides  K  0,  0  (x  and  the  included  angle 
in  the  one  equal  to  the  two  sides  1 0,  0  H  and  tlie  includ- 
ed angle  in  the  other,  each  to  each ;  hence  the  angle 
K  G  0  is  equal  to  the  angle  I H  0  (Prop.  V.  Cor.).  But, 
by  hypothesis,  the  angle  KGO  is  equal  to  the  angle  CHO, 
therefore  the  angle  1 H  0  is  equal  to  CHO,  so  that  H  I 
and  H  C  must  coincide ;  that  is,  the  line  C  D  when  pro- 
duced meets  IK  in  two  points,  I,  K,  and  yet  does  not  form 
one  and  the  same  straight  line,  which  is  impossible  (Prop. 
III.)  ;  therefore  the  lines  A B,  CD  cannot  meet,  conse- 
quently they  are  parallel  (Art.  17). 

Note.  —  The  demonstration  of  the  proposition  is  substantially  that 
given  by  M.  da  Cunha  in  the  Prmcipes  Malhematiques.  This  demon- 
stration Young  pronounces  "  superior  to  every  other  that  has  been  given 
of  the  same  proposition  " ;  and  Professor  Playfair,  in  the  Edinburgh  lie- 
view,  Vol.  XX.,  calls  attention  to  it,  as  a  most  important  improvement 
in  elementary  Geometry. 

Peoposition  XXI.  —  Theorem. 

84.  If  a  straight  line,  intersecting  tivo  other  straight 
lines,  makes  any  exterior  angle  equal  to  the  interior  and 
opposite  angle,  or  makes  the  interior  angles  on  the  same 
side  together  equal  to  tiuo  right  angles,  the  two  lines  are 
parallel. 

Let  the  straight  line  E  F  inter-  E 

sect  the  two  straight  lines  A  B,  \ 

C  D,  making  the  exterior  angle   ^ yr, 

E  Gr  B  equal  to  the  interior  and 
opposite  angle,  G  H  D  ;  then  the 
lines  A B,  CD  are  parallel.  ^ 1^^-  d 

For  the  angle  A  G  H  is  equal  \ 

to  the  angle  E  G  B  (Prop.  IV.)  ;  F 

and  E  G  B  is  equal  to  G  H  D,  by  hypothesis  ;  therefore  the 
angle  A  G II  is  equal  to  the  angle  G  H  D  ;  and  they  are 
alternate  angles ;  hence  the  lines  A  B,  CD  are  parallel 
(Prop.  XX.). 


32  ELEMENTS    OF   GEOMETRY. 

Again,  let  the  interior  angles 
on  the  same  side,  B  GH,  GH  D, 
be  together  equal  to  two  right 
angles ;  then  the  lines  A  B,  C  D 
are  parallel. 

For  the  sum  of  the  angles 
BGH,  GHD  is  equal  to  two 
right  angles,  by  hypothesis  ;  and  F 

the  sum  of  AG H,  BGH  is  also  equal  to  two  right  an- 
gles (Prop.  I.)  ;  take  away  BGH,  which  is  common  to 
both,  and  there  remains  the  angle  GHD,  equal  to  the 
angle  A  G  H  ;  and  these  are  alternate  angles  ;  hence  the 
lines  AB,  CD  are  parallel. 


E 

B 


D 


85.    Cor,  If  two  straight  lines     A- 
are  perpendicular  to  another,  they 
are  parallel ;  thus  AB,  CD,  per- 
pendicular to  E  F,  are  parallel.     * 


Proposition  XXH. — Theorem. 

86.  If  a  straight  line  intersects  two  parallel  lines,  it 
viakes  the  alternate  angles  equal;  also  any  exterior  angle 
equal  to  the  interior  and  opposite  angle  ;  and  the  tivo  in- 
terior angles  upon  the  same  side  together  equal  to  two 
right  angles. 


Let  the  straight  line  E  F  inter- 
sect the  parallel  lines  AB,  CD  ; 
the  alternate  angles  AGH,  GHD 
are  equal ;  the  exterior  angle 
E  G  B  is  equal  to  the  interior  and 

opposite   angle   GHD;    and  the      q l\ d 

two  interior  angles  BGH,  GHD 
upon  the  same  side  are  together 
equal  to  two  right  angles. 


L 

A Y-|-- r-B 

K 


Ji^ 


BOOK  I.  33 

For  if  the  angle  A  G  H  is  not  equal  to  G  H  D,  draw  the 
straiglit  line  K  L  through  the  point  G,  making  the  angle 
K  G  H  equal  to  G  H  D  ;  then,  since  the  alternate  angles 
GHD,  KGH  are  equal,  KL  is  parallel  to  CD  (Prop. 
XX.)  ;  but  by  hypothesis  A  B  is  also  parallel  to  C  D,  so 
that  through  the  same  point,  G,  two  straight  lines  are 
drawn  parallel  to  C  D,  which  is  impossible  (Art.  34,  Ax. 
12).  Hence  the  angles  A  G  H,  GHD  are  not  unequal ; 
that  is,  they  are  equal. 

Now,  the  angle  E  G  B  is  equal  to  the  angle  A  G  H 
(Prop.  IV.)?  ^i^d  AGH  has  been  shown  to  be  equal  to 
GHD;  hence  E  G  B  is  also  equal  to  G  H  D. 

Again,  add  to  each  of  these  equals  the  angle  B  G  H ; 
then  the  sum  of  tlie  angles  E  G  B,  B  G  H  is  equal  to  the 
sum  of  the  angles  B  G  H,  G  H  D.  But  E  G  B,  B  G  H  are 
equal  to  two  right  angles  (Prop.  I.)  ;  hence  B  G  H, 
GHD  are  also  equal  to  two  right  angles. 

87.  Cor.  If  a  line  is  perpendicular  to  one  of  two 
parallel  lines,  it  is  perpendicular  to  the  other ;  thus 
EF  (Art.  85),  perpendicular  to  A  B,  is  perpendicular 
to   CD. 

Proposition  XXIII.  —  Theorem. 

88.  If  tvw  straight  lines  intersect  a  third  line,  and 
make  the  two  interior  angles  on  the  same  side  together 
less  than  tivo  right  angles,  the  tivo  lines  ivill  meet  on 
being  j)roduced. 

Let  the  two  lines  KL,  CD  make  j. 

with  EF  the  angles  KGH,  GHC, 
togetlicr  less  than  two  right  angles ; 
then  KL  and  CD  will  meet  on 
being  produced. 

For  if  tliey  do  not  meet,  they 
are  parallel  (Art.  17).  But  they 
are  not  parallel ;  for  then  the  sum 


34 


ELEMENTS    OF   GEOMETRY. 


of  tlie  interior  angles  K  G  IT,  G II  C  would  be  equal  to 
two  right  angles  (Prop.  XXII.)  ;  bat  by  hypothesis  it 
is  less  ;  therefore  the  lines  K  L,  CD  will  meet  on  being- 
prod  u  cod. 

89.  Scholium.  The  two  lines  K  L,  CD,  on  being  pro- 
duced, must  meet  on  the  side  of  E  F,  on  which  are  the  two 
interior  angles  whose  sum  is  less  than  two  right  angles. 


E 


Proposition  XXIY.  —  Theorem. 

90.   Slraig-M  lines  which  are  parallel  to  the  same  line 
are  parallel  to  each  other. 

Let  the  straight  lines  A  B,  C  D 
be  each  parallel  to  the  line  E  F  ; 
then  are  they  parallel  to  each  other. 

Draw  GHI  perpendicular  to 
EF.  Then,  since  AB  is  parallel 
to  EF,  Gl  will  be  perpendicular 
to  AB  (Prop.  XXII.  Cor.)  ;  and 
since  C  D  is  parallel  to  EF,  GI 
will  for  a  like  reason  be  perpendicular  to  CD. 


n 


D 


B 


straight  line  ; 


Conse- 
the   same 
hence  they  are  parallel  (Prop.  XXI.  Cor.). 


quently   A  B   and    C  D    are   perpendicular   to 


H 


G 


Proposition  XXV.  —  Theorem. 

91.    Tvw  parallel  straight  lines  are  everywhere  equallij 
distant  from  each  other. 

Let  AB,  CD  be  two  parallel 
straight  lines.  Through  any  two 
points  in  A  B,  as  E  and  F,  draw 
the  straight  lines  E  G,  F  H,  per- 
pendicular to  A  B.  These  lines 
will  be  equal  to  each  other. 

For,  if  GF  be  joined,  the  angles  GF  E,  F  G  H,  consid- 
ered in  reference  to  the  parallels  AB,  CD,  will  be  alter- 


E 


D 


B 


BOOK   I.  85 

nate  interior  angles,  and  tlierefore  equal  to  each  other 
(Prop.  XXII.).  Also,  since  the  straight  lines  E  G,  F  H 
are  perpendicular  to  the  same  straiglit  line  AB,  and  con- 
sequently parallel  (Prop.  XXI.  Cor.),  the  angles  EGF, 
GFli,  considered  in  reference  to  the  parallels  EG,  FH, 
will  be  alternate  interior  angles,  and  therefore  equal. 
Hence,  the  two  triangles  E  F  G,  F  G  H,  have  a  side  and 
the  two  adjacent  angles  of  the  one  equal  to  a  side  and  the 
two  adjacent  angles  of  the  other,  eacli  to  each ;  therefore 
these  triangles  are  equal  (Prop.  VI.)  ;  hence  the  side  E  G, 
which  measures  the  distance  of  the  parallels  A B,  CD, 
at  the  point  E,  is  equal  to  the  side  F  H,  which  measures 
the  distance  of  the  same  parallels  at  the  point  F.  Hence 
two  parallels  are  everywhere  equally  distant. 

Proposition  XXVI.  —  Theorem. 

92.  If  Uvo  angles  have  their  sides  parallel,  each  to 
each,  and  lying^  in  the  same  direction,  the  two  angles  are 
equal. 

Let  A  B  C,  D  E  F  be  two  angles,  ^        j^ 

which  have  the  side  AB  parallel  /         / 

to  HE,  and  BC  parallel  to  EF;  /   ^/ ^ 

then  these  angles  are  equal.  /       7 

For  produce  HE,  if  necessary,  ^        7^  CJ 

till  it  meets  B  C  in  the  point  G.     h ^ F 

Then,  since  E  F  is  parallel  to  G  C, 

the  angle  D  E  F  is  equal  to  D  G  C  (Prop.  XXII.)  ;  and 
since  D  G  is  parallel  to  A  B,  the  angle  H  G  C  is  equal 
to  ABC;  hence  the  angle  H  E  F  is  equal  to  ABC. 

93.  Scholiu7n.  This  proposition  is  restricted  to  the  case 
where  the  side  E  F  lies  in  the  same  direction  with  B  C, 
since  if  F  E  were  produced  toward  H,  the  angles  D  E  H, 
ABC  would  only  be  equal  when  they  are  right  angles. 


36 


ELEMENTS   OF   GEOMETRY. 


Proposition  XXYII.  —  Theorem. 

94.  If  any  side  of  a  triang-Ie  be  produced^  the  exterior 
ang-le  is  equal  to  the  sum  of  the  tivo  interior  and  opposite 
angles. 

Let  ABC  be  a  triangle,  and 
let  one  of  its  sides,  B  C  be  pro- 
duced towards  D  ;  then  the  ex- 
terior angle  A  C  D  is  equal  to 
the  two  interior  and  opposite  g 
angles,  CAB,  ABC. 

For,  draw  E  C  parallel  to  the  side  AB  ;  then,  since  AC 
meets  the  two  parallels  A  B,  EC,  the  alternate  angles 
B  AC,  A  C  E  are  equal  (Prop.  XXII.). 

Again,  since  BD  meets  the  two  parallels  AB,  EC,  the 
exterior  angle  E  C  D  is  equal  to  the  interior  and  opposite 
angle  ABC.  But  the  angle  ACE  is  equal  to  BAC; 
therefore,  the  whole  exterior  angle  A  C  D  is  equal  to  the 
two  interior  and  opposite  angles  CAB,  ABC  (Art.  34, 
Ax.  2). 


Proposition  XXYIII.  —  Theorem. 

95.  In  every  triangle  the  sum  of  the  three  angles  is 
equal  to  two  right  angles. 

Let  A  B  C  be  any  triangle  ; 
then  will  the  sum  of  the  angles 
ABC,  BCA,  CAB  be  equal 
to  two  right  angles. 

For,  let  the  side  B  C  be  pro-  g 
duced  towards  D,  making  the 
exterior  angle  A  C  D  ;  then  the  angle  A  C  D  is  equal  to 
CAB  and  ABC  (Prop.  XXVII.).  To  each  of  these 
equals  add  the  angle  A  C  B,  and  we  shall  have  the  sum  of 


BOOK    I.  37 

A  C  B  and  A  C  D,  equal  to  the  sum  of  A  B  C,  B  C  A,  and 
CAB.  But  the  sum  of  A  C  B  and  A  C  D  is  equal  to  two 
right  angles  (Prop.  I.)  ;  hence  the  sum  of  the  three  an- 
gles ABC,  B  C  A,  and  C  A  B  is  equal  to  two  right  angles 
(Art.  34,  Ax.  2). 

96.  Cor.  1.  Two  angles  of  a  triangle  being  given,  or 
merely  their  sum,  the  third  will  be  found  by  subtracting 
that  sum  from  two  right  angles. 

97.  Cor.  2.  If  two  angles  in  one  triangle  be  respective- 
ly equal  to  two  angles  in  another,  their  third  angles  will 
also  be  equal. 

98.  Cor.  3.  A  triangle  cannot  have  more  than  one  an- 
gle as  great  as  a  right  angle. 

99.  Cor.  4.  And,  therefore,  every  triangle  must  have 
at  least  two  acute  angles. 

100.  Cor.  5.  In  a  right-angled  triangle  the  right  angle 
is  equal  to  tlie  sum  of  the  other  two  angles. 

101.  Cor.  6.  Since  every  equilateral  triangle  is  also 
equiangular  (Prop.  VII.  Cor.  3),  each  of  its  angles  will 
be  equal  to  two  thirds  of  one  right  angle. 

Proposition  XXIX.  —  Theorem. 

102.  The  sum  of  all  the  interior  angles  of  any  polygon 
is  equal  to  tivice  as  many  right  angles,  less  four,  as  the 
figure  has  sides. 

Let  A  B  C  D  E  be  any  polygon  ;  then 
the  sum  of  all  its  interior  angles.  A,  B, 
C,  D,  E,  is  equal  to  twice  as  many 
right  angles  as  the  figure  has  sides,  less 
four  right  angles. 

For,  from  any  point  P  within  the  pol-  A  B 

ygon,  draw  tlie  straight  lines  PA,  PB,  PC,  P  D,  P  E,  to 
the  vertices  of  all  the  angles,  and  the  polygon  will  be 

4 


38  ELEMENTS   OF   GEOMETRY. 

divided  into  as  many  triangles  as  it  has  sides.  Now,  the 
sum  of  the  three  angles  in  each  of  these  triangles  is  equal 
to  two  right  angles  (Prop.  XXVIII.)  ;  therefore  the  sum 
of  the  angles  of  all  these  triangles  is  equal  to  twice  as 
many  right  angles  as  there  are  triangles,  or  sides,  to  the 
polygon.  But  the  sum  of  all  the  angles  about  the  point 
P  is  equal  to  four  right  angles  (Prop.  TV.  Cor.  2),  which 
sum  forms  no  part  of  the  interior  angles  of  the  polygon  ; 
therefore,  deducting  the  sum  of  the  angles  about  the  point, 
there  remain  the  angles  of  the  polygon  equal  to  twice  as 
many  right  angles  as  the  figure  has  sides,  less  four  right 
angles. 

103.  Cor.  1.  The  sum  of  the  angles  in  a  quadrilateral 
is  equal  to  four  right  angles  ;  hence,  if  all  the  angles  of  a 
quadrilateral  are  equal,  each  of  them  is  a  right  angle  ; 
alsoy  if  three  of  the  angles  are  right  angles,  the  fourth  is 
likewise  a  right  angle. 

104.  Cor.  2.  The  sum  of  the  angles  in  a  pentagon  is 
equal  to  six  right  angles  ;  in  a  hexagon^  the  sum  is  equal 
to  eight  right  angles,  &c. 

105.  Cor.  3.  In  every  equiangular  figure  of  more  than 
four  sides,  each  angle  is  greater  than  a  right  angle  ;  thus, 
in  a  regular  pentagon^  each  angle  is  equal  to  one  and  one 
fifth  right  angles  ;  in  a  regular  hexagon^  to  one  and  one 
third  right  angles,  <fec. 

106.  Scholium.  In  applying  this  prop-  d 
osition  to  polygons  which  have  re-en- 
trant angles,  or  angles  whose  vertices 
are  directed  inward,  as  BPC,  each  of 
these  angles  must  be  considered  greater 
than  two  right  angles.     But,  in  order 
to  avoid  ambiguity,  we  shall  hereafter 
limit  our  reasoning  to   polygons  with  salient  angles,  or 
with  angles  directed  outwards,  and  which  may  be  called 
convex  polygons.     Every  convex  polygon  is  such  that  a 


BOOK   I.  39 

straight  line,  however  drawn,  cannot  meet  the  perimeter 
of  the  polygon  in  more  than  two  points. 

Proposition  XXX.  —  Theorem. 

107.  The  sutn  of  all  the  ex^rior  angles  of  any  pohjgon^ 
formed  hy  producing  each  side  in  the  same  direction^  is 
equal  to  four  right  angles. 

Let  each  side  of  the  polygon  ABODE 
he  produced  in  the  same  direction ;  then 
the  sum  of  the  exterior  angles  A,  B,  C, 
D,  E,  will  be  equal  to  four  right  angles. 

For  each  interior  angle,  together  with 
its  adjacent  exterior  angle,  is  equal  to 
two  right  angles  (Prop.  1.)  ;  hence  the  sum  of  all  tlie 
angles,  both  interior  and  exterior,  is  equal  to  twice  as 
many  right  angles  as  there  are  sides  to  the  polygon.  But 
the  sum  of  the  interior  angles  alone,  less  four  right  angles, 
is  equal  to  the  same  sum  (Prop.  XXIX.)  ;  therefore  the 
sum  of  the  exterior  angles  is  equal  to  four  right  angles. 

Proposition  XXXI.  —  Theorem. 

108.  The  opposite  sides  and  angles  of  every  parallelo- 
gram are  equal  to  each  other. 

Let  A  B  C  D  be  a  parallelogram ;  p)  C 

then  the  opposite  sides  and  angles  are         y^x  7 

equal  to  each  other.  /  ...         / 

Draw  the  diagonal  B  D,  then,  since     /_ x  / 

the  opposite  sides  AB,  DC  are  paral-    A  B 

lei,  and  BD  meets  them,  the  alternate  angles  ABD,  BDC 
are  equal  (Prop.  XXII.)  ;  and  since  AD,  BC  are  parallel, 
and  B  D  meets  them,  the  alternate  angles  A  D  B,  D  B  C 
are  likewise  equal.  Hence,  the  two  triangles  AD  B,  D  B  C 
liave  two  angles,  A  B  D,  A  D  B,  in  tlie  one,  equal  to  two 
angles,  BDC,  DBC,  in  the  otlier,  eacli  to  each;  and  since 


40  ELEMENTS   OF   GEOMETRY. 

the  side  BD  included  between  these 
equal  angles  is  common  to  the  two  tri- 
angles, they  are  equal  (Prop.  VI.)  ; 
hence  the  side  AB  opposite  the  angle 
A  D  B  is  equal  to  the  side  D  C  opposite 
the  angle  DBG  (Prop.  Yl.  Cor.)  ;  and,  in  like  manner, 
the  side  A  D  is  equal  to  the  side  B  C ;  hence  the  opposite 
sides  of  a  parallelogram  are  equal. 

Again,  since  the  triangles  are  equal,  the  angle  A  is  equal 
to  the  angle  C  (Prop.  VI.  Cor.)  ;  and  since  the  two  angles 
D  B  C,  A  B  D  are  respectively  equal  to  the  two  angles 
A  D  B,  B  D  C,  the  angle  ABC  is  equal  to  the  angle 
ADC. 

109.  Cor.  1.  The  diagonal  divides  a  parallelogram  into 
two  equal  triangles. 

110.  Cor.  2.  The  two  parallels  AD,  B  C,  included  be- 
tween two  other  parallels,  AB,  CD,  are  equal. 

Proposition  XXXII. — Theorem. 

111.  If  the  opposite  sides  of  a  quadrilateral  are  equal, 
each  to  each,  the  equal  sides  are  parallel,  and  the  figure 
is  a  parallelogram. 

Let    A  B  C  D    be   a   quadrilateral          D  C 

having  its  opposite  sides  equal ;  then            /\  7 

will  the  equal  sides  be  parallel,  and          /        \        / 
the  figure  be  a  parallelogram.  / jj  / 

For,   having   drawn   the    diagonal        ^  ^ 

B  D,  the  triangles  A  B  D,  B  D  C  have  all  the  sides  of  tlie 
one  equal  to  the  corresponding  sides  of  the  other  ;  there- 
fore they  are  equal,  and  the  angle  A  D  B  opposite  the  side 
A  B  is  equal  to  D  B  C  opposite  C  D  (Prop.  XVIII.  Sch.)  ; 
hence  the  side  A  D  is  parallel  to  B  C  (Prop.  XX.).  For 
a  like  reason,  AB  is  parallel  to  C  D  ;  therefore  the  quad- 
rilateral A  B  C  D  is  a  parallelogram. 


BOOK   I.  41 


Proposition  XXXIII.  —  Theorem. 

112.  If  two  opposite  sides  of  a  quadrilateral  are  equal 
and  parallel,  the  other  sides  are  also  equal  and  parallel, 
and  the  figure  is  a  parallelogram. 

Let   A  B  C  D    be   a  quadrilateral,          D  C 

having  the  sides  A B,  CD  equal  and        ^\  7 

parallel ;   then  will   the  other   sides       /  / 

also  be  equal  and  parallel.  [_ '''■■■■.  / 

Draw  the  diagonal  BD ;  then,  since    ^  ^ 

A  B  is  parallel  to  C  D,  and  B  D  meets  them,  tlie  alternate 
angles  ABD,  BD  C  are  equal  (Prop.  XXII.)  ;  moreover, 
in  the  two  triangles  ABD,  D  B  C,  the  side  B  D  is  com- 
mon ;  therefore,  two  sides  and  the  included  angle  in  the 
one  are  equal  to  two  sides  and  the  included  angle  in  the 
other,  each  to  each ;  hence  these  triangles  are  equal 
(Prop.  Y.),  and  the  side  A  D  is  equal  to  B  C.  Hence 
the  angle  A  D.B  is  equal  to  D  B  C,  and  consequently  A  D 
is  parallel  to  B  C  (Prop.  XX.)  ;  therefore  the  figure 
A  B  C  D  is  a  parallelogram. 

Proposition  XXXIV.  —  Theorem. 

113.    The  diagonals  of  every  parallelogram  bisect  each 
other. 

Let   A  B  C  D   be  a  parallelogram,  D  C 

and  AC,  DB  its  diao'onals,  intersect-  /x 

ing  at  E  ;  then  will  A E  equal  EC,  /       .'.:.-:;.;' 

and  B  E  equal  E  D.  /■•--••"""      "'•-• 

For,  since  A  B,  C  D  are  parallel,  A 
and  B  D  meets  them,  the  alternate  angles  C  D  E,  ABE 
are  equal  (Prop.  XXII.)  ;  and  since  A  C  meets  the  same 
parallels,  tlie  alternate  angles  BAE,  ECD  are  also  equal; 
and  the  sides  AB,  CD  arc  equal  (Prop.  XXXI.).  Hence 
the  triangles  ABE,  CDE  have  two  angles  and  tlie  in- 

4* 


42  ELEMENTS   OF   GEOMETRY. 

eluded  side  in  the  one  equal  to  two  angles  and  the  includ- 
ed side  in  the  other,  each  to  each ;  hence  the  two  triangles 
are  equal  (Prop.  VI.)  ;  therefore  the  side  A  E  opposite  the 
angle  A  B  E  is  equal  to  C  E  opposite  C  D  E  ;  hence,  also, 
the  sides  B  E,  D  E  opposite  the  other  equal  angles  are 
equal. 

114.  Scholium.  In  the  case  of  a  rhom- 
bus, the  sides  A  B,  B  C  being  equal,  the 
triangles  A  E  B,  E  B  C  have  all  the  sides 
of  tlie  one  equal  to  the  corresponding 
sides  of  the  other,  and  are,  therefore, 
equal ;  whence  it  follows  that  the  angles  A  E  B,  BEG 
are  equal.  Therefore  the  diagonals  of  a  rhombus  bisect 
each  other  at  right  angles. 

Peoposition  XXXY.  —  Theorem. 

115.  If  the  diagonals  of  a  quadrilateral  bisect  each 
other ^  the  figure  is  a  parallelogram. 

Let  A  B  C  D  be  a  quadrilateral,  and     D  C 

AC,  D  B  its  diagonals  intersecting  at  E  ;      /\  .■■•■-""/ 

then  will  the  figure  be  a  parallelogram.        /      j'>C       / 

For,  in  the  two  triangles  ABE,  CDE,  /•••••-•""         '''•'••.../ 


the  two  sides  A  E,  E  B  and  the  hicluded  ^  ^ 

angle  in  the  one  are  equal  to  the  two  sides  C  E,  E  D  and 
the  included  angle  in  the  other ;  hence  the  triangles  are 
equal,  and  the  side  AB  is  equal  to  the  side  CD  (Prop.  Y. 
Cor.).  For  a  like  reason,  A  D  is  equal  to  C  B  ;  therefore 
the  quadrilateral  is  a  parallelogram  (Prop.  XXXII.). 


BOOK    II. 

RATIO    AND    PROPORTION. 
DEFINITIONS. 

116.  Ratio  is  the  relation,  in  respect  to  quantity,  which 
one  magnitude  bears  to  another  of  tlie  same  kind  ;  and  is 
tlie  quotient  arising  from  dividing  the  first  by  the  second. 

A  ratio  may  be  written  in  the  form  of  a  fraction,  or 
witli  the  sign  :  . 

Thus  the  ratio  of  A  to  B  may  be  expressed  either  by 

g,  or  by  A  :  B. 

IIT.  The  two  magnitudes  necessary  to  form  a  ratio  are 
called  the  terms  of  tlie  ratio.  The  first  term  is  called  the 
ANTECEDENT,  and  tlic  last,  the  consequent. 

118.  Ratios  of  magnitudes  may  be  expressed  by  num- 
bers, either  exactly,  or  approximately. 

This  may  be  illustrated  by  the  operation  of  finding  the 
numerical  ratio  of  two  straight  lines,  AB,  CD. 

From  the  crreater  line  ^ 

O  Q, , ,jy 

AB  cut  off  a  part  equal  ^ 

to  the  less  C  D,  as  many 

times  as  possible  ;  for  ex-  E      G 

ample,  twice,  with  the  remainder  B  E. 

From  the  line  C  D  cut  off  a  part  equal  to  the  remainder 
B  E  as  many  times  as  possible ;  once,  for  example,  with 
the  remainder  D  F. 

From  the  first  remainder  B  E,  cut  off  a  part  equal  to 
the  second  D  F,  as  many  times  as  possible  ;  once,  for  ex- 
ample, with  tlie  remainder  B  G. 


44  ELEMENTS    OF   GEOMETRY. 

-• ^D 


A*- 


From  the  second  re- 
mainder DF,  cut  off  a 
part   equal    to    B  G,   the 

thh^d,  as  many   times  as      ^'  '  15      G~ 

possible. 

Proceed  tlms  till  a  remainder  arises,  which  is  exactly 
contained  a  certain  number  of  times  in  the  preceding  one. 

Then  this  last  remainder  will  be  the  common  measure 
of  the  proposed  lines  ;  and,  regarding  it  as  unity,  we  sliall 
easily  find  the  values  of  the  preceding  remainders  ;  and, 
at  last,  those  of  the  two  proposed  lines,  and  hence  their 
ratio  in  numbers. 

Suppose,  for  instance,  we  find  G  B  to  be  contained  ex- 
actly twice  in  F  D  ;  B  G  will  be  the  common  measure  of 
the  two  proposed  lines.  Let  B  G  equal  1 ;  then  will  F  D 
equal  2..  But  EB  contains  FD  once,  plus  GB;  there- 
fore we  have  E B  equal  to  3.  CD  contains  E  B  once, 
plus  FD  ;  therefore  we  have  CD  equal  to  5.  AB  con- 
tains C  D  twice,  plus  E  B  ;  therefore  we  have  A  B  equal 
to  13.  Hence  the  ratio  of  the  two  lines  is  that  of  13  to  5. 
If  the  line  C  D  were  taken  for  unity,  the  line  A  B  would 
be  V- ;  if  A  B  were  taken  for  unity,  C  D  would  be  i\. 

It  is  possible  that,  however  far  the  operation  be  con- 
tinued, no  remainder  may  be  found  which  shall  be  con- 
tained an  exact  number  of  times  in  the  preceding  one. 
In  that  case  there  can  be  obtained  only  an  approximate 
ratio,  expressed  in  numbers,  more  or  less  exact,  according 
as  the  operation  is  more  or  less  extended. 

119.  When  the  greater  of  two  magnitudes  contains  the 
less  a  certain  number  of  times  without  having  a  remain- 
der, it  is  called  a  multiple  of  the  less  ;  and  the  less  is  then 
called  a  submultiple,  or  measure  of  the  greater. 

Thus,  6  is  a  multiple  of  2  ;  2  and  3  are  submultiples, 
or  measures,  of  6. 

120.  Equimultiples,  or  like  multiples,  are  those  which 
contain  their  respective  submultiples  the  same  number  of 


BOOK  II.  45 

times  ;  and  equisubmultiples,  or  like  submultiples,  are 
tliose  contained  in  their  respective  multiples  the  same 
number  of  times. 

Thus  4  and  5  arc  like  submultiples  of  8  and  10  ;  8  and 
10  are  like  multiples  of  4  and  5. 

121.  Commensurable  magnitudes  are  magnitudes  of 
the  same  kind,  which  have  a  common  measure,  and  whose 
ratio  therefore  may  be  exactly  expressed  in  numbers. 

122.  Incommensurable  magnitudes  are  magnitudes  of 
the  same  kind,  which  have  no  common  measure,  and 
whose  ratio,  therefore,  cannot  be  exactly  expressed  m 
numbers. 

123.  A  direct  ratio  is  the  quotient  of  the  antecedent  by 
the  consequent ;  an  inverse  ratio,  or  recipIiocal  ratio,  is 
the  quotient  of  the  consequent  by  the  antecedent,  or  the 
reciprocal  of  the  direct  ratio. 

Thus  the  direct  ratio  of  a  line  6  feet  long  to  a  line  2  feet 
long  is  I  or  3  ;  and  the  inverse  ratio  of  a  line  6  feet  long 
to  a  line  2  feet  long  is  |  or  ~,  which  is  the  same  as  the 
reciprocal  of  3,  the  direct  ratio  of  6  to  2. 

The  word  ratio  when  used  alone  means  the  direct  ratio. 

124.  A  compound  ratio  is  the  product  of  two  or  more 
ratios. 

Thus   the   ratio   compounded   of  A  :  B  and    C  :  D    is 
A        C         A  X  C 
B  -^  5'  ^''  B  X  D* 

125.  A  proportion  is  an  equality  of  ratios. 

Four  magnitudes  are  in  proportion,  when  the  ratio  of 
the  first  to  the  second  is  the  same  as  that  of  the  third  to 
the  fourth. 

Thus,  the  ratios  of  A  :  B  and  X  :  Y,  being  equal  to 

A       X 
each  other,  when  written   A  :  B  =  X  :  Y,  or  —  =  y) 

form  a  proportion. 


46  ELEMENTS    OF   GEOMETRY. 

126.  Proportion  is  written  not  only  with  the  sign  =, 
but,  more  often,  with  the  sign  : :  between  the  ratios. 

Thus,  A  :  B  : :  X  :  Y,  expresses  a  proportion,  and  is 
read.  The  ratio  of  A  to  B  is  equal  to  the  ratio  of  X  to  Y ; 
or,  A  is  to  B  as  X  is  to  Y. 

-  127.,  TliQ  first  and  third  terms  of  a  proportion  are  called 
the  antecedents;  the  second  and  fourth,  the  consequents. 
T\iQ  first  mid  fourth  are  also  called  the  extremes,  and  the 
second  and  third  the  means. 

Thus,  in  the  proportion  A  :  B  : :  C  :  D,  A  and  C  are 
the  antecedents ;  B  and  D  are  the  consequents  ;  A  and  D 
are  the  extremes ;  and  B  and  C  are  the  means. 

The  antecedents  are  called  homologous  or  like  terms, 
and  so  also  are  the  consequents. 

128.  All  the  terms  of  a  proportion  are  called  propor- 
tionals ;  and  the  last  term  is  called  a  fourth  propor- 
tional to  the  other  three  taken  in  their  order. 

Thus,  in  the  proportion  A  :  B : :  C  :  D,  D  is  the  fourth 
proportional  to  A,  B,  and  C. 

129.  When  both  the  means  are  the  same  magnitude, 
either  of  them  is  called  a  mean  proportional  between  the 
-extremes  ;  and  if,  in  a  series  of  proportional  magnitudes, 
each  consequent  is  the  same  as  the  next  antecedent,  those 
magnitudes  are  said  to  be  in  continued  proportion. 

Thus,  if  we  have  A  :  B  : :  B  :  C  :  :  C  :  D  :  :  D  :  E,  B  is  a 
mean  proportional  between  A  and  C,  C  between  B  and  D, 
J)  between  C  and  E  ;  and  the  magnitudes  A,  B,  C,  D,  E 
are  said  to  be  in  continued  proportion. 

130.  When  a  continued  proportion  consists  of  but  three 
terms,  the  middle  term  is  said  to  be  a  mean  proportional 
between  the  other  two ;  and  the  last  term  is  said  to  be  the 
third  proportional  to  tlie  first  and  second. 

Thus,  when  A,  B,  and  C  are  in  proportion,  A :  B  : :  B  :  C ; 
in  which  case  B  is  called  a  moan  proportional  between  A 
and  C  ;  and  C  is  called  the  third  proportional  to  A  and  B. 


BOOK    II.  47 

131.  Magnitudes  are  in  proportion  by  inversion,  or 
INVERSELY,  when  each  antecedent  takes  the  place  of  its 
consequent,  and  each  consequent  the  place  of  its  antece- 
dent. 

Thus,  let  A  :  B  :  :  C  :  D  ;  then,  by  inversion, 
B  :  A  :  :  D  :  C. 

132.  Magnitudes  are  in  proportion  by  alternation,  or 
ALTERNATELY,  whcu  antecedent  is  compared  with  antece- 
dent, and  consequent  with  consequent. 

Thus,  let  A  ;  B  :  :  D  :  C  ;  then,  by  alternation, 
A  :  D  : :  B  :  C. 

133.  Magnitudes  are  in  proportion  by  composition, 
when  the  sum  of  the  first  antecedent  and  consequent  is 
to  the  first  antecedent,  or  consequent,  as  the  sum  of  the 
second  antecedent  and  consequent  is  to  the  second  ante- 
cedent, or  consequent. 

Thus,  let  A  :  B  :  :  C  :  D  ;  then,  by  composition, 
A  +  B  :  A  : :  C  +  D  :  C,    or     A  +  B  :  B  : :  C  +  D  :  D. 

134.  Magnitudes  are  in  proportion  by  division,  when 
the  difference  of  the  first  antecedent  and  consequent  is  to 
the  first  antecedeiit,  or  consequent,  as  the  difference  of  the 
second  antecedent  and  consequent  is  to  the  second  ante- 
cedent, or  consequent. 

Thus,  let  A  :  B  :  :  C  :  D  ;  then,  by  division, 
A  — B:A::C  — D:C,    or   A  — B  :  B  :  :  C  — D  :  D. 

Proposition  I.  —  Theorem. 

135.  If  four  magnitudes  are  in  proportion^  the  product 
of  the  tvjo  extremes  is  equal  to  the  product  of  the  tivo 
means. 

Let  A  :  B  :  :  C  :  D  ;  then  will  A  X  I)  =  B  ^^^^^^^^ 
For,  since  the  magnitudes  arc  in  proportioj^^^  <^ 


B  =  D'        /f%iV*  v*:^: 


48  ELEMENTS   OF    GEOMETRY. 

and  reducing  the  fractions  of  this  equation  to  a  common 
denominator,  we  have 

AX  D  ^  B  X  C 
B  X  i)         B  X  D' 
or,  the  common  denominator  being  omitted, 
AX  D  =  B  X  C. 

Proposition  II.  —  Theorem. 

136.  If  the  product  of  two  magnitudes  is  equal  to  the 
product  of  two  others^  these  four  magnitudes  form  a 
proportion. 

Let  A  X  D  =  B  X  C  ;  then  will  A  :  B  : :  C  :  D. 
For,  dividing  each  member  of  the  given  equation  by 
B  X  D,  we  have 

AX  D  _  BX  C 
B  X  D         B  X  D' 
which,  reduced  to  the  lowest  terms,  gives 
A  ^  0 
B        D' 
Whence  A  :  B  : :  0  :  D. 

Proposition  III. — Theorem. 

137.  If  three  magnitudes  are  in  proportion^  the  product 
of  the  two  extremes  is  equal  to  the  square  of  the  mean. 

Let  A  :  B  :  :  B  :  C ;  then  will  A  X  C  =  B^ 
For,  since  the  magnitudes  are  in  proportion, 
A  ^  B 
B        C' 
and,  by  Prop.  L, 

AXC:=-BXB,     or    A  X  C  =  B^ 


BOOK   II.  49 


Proposition  IV .  —  Theorem. 

138.  If  the  product  of  any  tvw  quantities  is  equal  to 
the  square  of  a  third,  the  third  is  a  mean  proportional 
between  the  other  two. 

Let  Ax  C  =:  B^ ;  then  B  is  a  mean  proportional  be- 
tween A  and  C. 
For,  dividing  each  member  of  the  given  equation  hj 

B  X  C,  we  have 

A  ^  B 

B  ~  C' 

whence  A  :  B  : :  B  :  C. 

Proposition  Y.  —  Theorem. 

139.  If  four  mag-nitudes  are  in  proportion,  they  will  be 
in  proportion  when  taken  inversely. 

Let  A  :  B  : :  C  :  D  ;  then  will  B  :  A  : :  D  :  C. 
For,  from  the  given  proportion,  by  Prop.  I.,  we  have 
AXD  =  BXC,     or    BxC  =  AxD. 
Hence,  by  Prop.  II., 

B  :  A  :  :  D  :  C. 

Proposition  VI.  —  Theorem. 

140.  If  four  magnitudes  are  in  proportion,  they  will  be 
in  proportion  when  taken  alternately. 

Let  A  :  B  : :  C  :  D ;  then  will  A  :  C  : :  B  :  D. 
For,  since  the  magnitudes  are  in  proportion, 
A         C 

B==D' 

B 

and  multiplying  each  member  of  this  equation  by  p,  we 

have 

A  X  B  _ 
B  X  C  ~ 


50  ELEMENTS    OF   GEOMETRY. 

which,  reduced  to  the  lowest  terms,  gives 

A  _  B 

C  ~  D' 
whence  A  :  C  : :  B  :  D. 

Proposition  VII.  —  Theorem. 

141.  If  four  magnittides  are  in  proportion^  they  will  be 
in  proportion  by  composition. 

Let  A :  B  :  :  C  :  D  ;  then  will  A  +  B  :  A :  :  C  +  D  :  C. 
For,  from  the  given  proportion,  by  Prop.  I.,  we  have 

BX  C  =  AX  D. 
Adding  A  X  C  to  each  side  of  this  equation,  we  have 

AXC  +  BXC  =  AXC  +  AXD, 
and  resolving  each  member  into  its  factors, 

(A  +  B)  X  C  =  (C  +  D)  X  A. 
Hence,  by  Prop.  II., 

A  +  B:A::C  +  D:C. 

Proposition  YIII.  —  Theorem. 

142.  If  four  magnitudes  are  in  proportion^  they  ivill  be 
in  proportion  by  division. 

Let  A  :  B  :  :  C  :  D  ;  then  will  A---  B  :  A  : :  C  —  D  :  C. 
For,  from  the  given  proportion,  by  Prop.  I.,  we  have 
B  X  C  =  A  X  D. 

Subtracting  each  side  of  this  equation  from  A  X  C,  we 

have 

AxC-^BxC  =  AxC  —  AxD, 

and  resolving  each  member  into  its  factors, 

(A  —  B)  X  C  =  (C  —  D)  X  A. 

Hence,  by  Prop.  II. , 

A  —  B  :  A  :  :  C  -^  D  :  C. 


BOOK   II.  51 


Proposition  IX.  —  Theorem. 

143.  Equimultiples  of  tivo  magnitudes  have  the  same 
ratio  as  the  magnitudes  themselves. 

Let  A  and  B  be  two  magnitudes,  and  m  X  A  and  m  X  B 
their  equimultiples,  then  will  m  X  A  :  m  X  B  :  ;  A :  B. 
For  A  X  B  =  B  X  A ; 

Multiplying  each  side  of  tins  equation  by  any  number, 
m,  we  have 

m  X  A  X  B  =  wi  X  B  X  A ; 
therefore 

Qn  X  A)  X  B  =  (wi  X  B)  X  A. 

Hence,  by  Prop.  II., 

w  X  A  :  wi  X  B  :  :  A  :  B. 

Proposition  X.  —  Theorem. 

144.  Magnitudes  which  are  proportional  to  the  same 
proportionals^  will  be  proportional  to  each  other. 

Let  A  :  B  :  :  E  :  F,  and  C  :  D  :  :  E  :  F ;  then  will 

A  :  B  :  :  C  :  D. 

For,  by  the  given  proportions,  we  have 

A         E  C         E 

-  =  -,  and   -  =  ^. 

Therefore,  it  is  evident  (Art.  34,  Ax.  1), 

A  __  C 

B  ~~  D* 
Hence  A  :  B  :  :  C  :  D. 

145.  Cor.  1.  If  two  proportions  have  an  antecedent 
and  its  consequent  the  same  in  both,  the  remaining  terms 
will  be  in  proportion. 

146.  Cor.  2.  Therefore,  by  alternation  (Prop.  VL),  if 
two  proportions  have  the  two  antecedents  or  the  two  con- 


62  ELEMENTS    OF   GEOMETRY. 

sequeiits  the  same  in  both,  the  remainmg  terms  will  be  in 
proportion. 

Proposition  XI.  —  Theorem. 

147.  If  any  number  of  magnitudes  are  proportional^ 
any  antecedent  is  to  its  consequent  as  the  sum  of  all  the 
antecedents  is  to  the  sum  of  all  the  consequents. 

Let  A  :  B  :  :  C  :  D  :  :  E  :  F ;  then  will 

A:B::A+C  +  E:BH-D  +  F. 
For,  from  the  given  proportion,  we  have 

A  X  D  =  B  X  C,      and      A  X  F  ==  B  X  E. 

By  adding  A  X  B  to  the  sum  of  the  corresponding  sides 
of  these  equations,  we  have 

AxB+AxD+AxF=AxB+BxC+Bxfi. 
Therefore, 

A  X  (B  +  D  +  F)  =  B  X  (A  +  C  +  E). 

Hence,  by  Prop.  II., 

A:B::A  +  C  +  E:B  +  D  +  F. 

Proposition  XII.  —  Theorem. 

148.  If  four  magnitudes  are  in  proportion^  the  sum  of 
the  first  and  second  is  to  their  difference  as  the  sum  of 
the  third  and  fourth  is  to  their  difference. 

Let  A  :  B  :  :  C  :  D ;  then  will 

A  +  B:A  —  B::C  +  D:C  —  D. 
For,  from  the  given  proportion,  by  Prop.  YII.,  we  have 
A  +  B:A::C  +  D:C; 
and  from  the  given  proportion,  by  Prop.  VIII.,  we  have 
A  — B:A::C  — D:C. 

Hence,  from  these  two  proportions,  by  Prop.  X.  Cor.  2, 

we  have 

A  +  B:A  —  B::C  +  D:C  —  D. 


BOOK   II.  63 


Proposition  XIII.  —  Theorem. 

149.  If  there  be  two  sets  of  proportional  magnitudes, 
the  products  of  the  corresponding  terms  will  be  propor- 
tionals. 

Let  A  :  B  :  :  C  :  D,  and  E  :  F  :  :  G  :  H ;  then  wiU 
AXE:BXF::CXG:DXH. 

For,  from  the  first  of  the  given  proportions,  by  Prop.  I., 
we  have 

A  X  D  =  B  X  C; 

and  from  the  second  of  the  given  proportions,  by  Prop.  I., 
we  have 

E  X  H  =  F  X  G. 

Multiplying  together  the  corresponding  members  of 
these  equations,  we  have 

AxDxExH  =  BxCxFxG. 

Hence,  by  Prop.  II., 

AxE:BxF::CxG:DxH. 

Proposition  XIY.  —  Theorem. 

150.  If  three  viagnitudes  are  proportionals,  the  first  will 
be  to  the  third  as  the  square  of  the  first  is  to  the  square 
of  the  second. 

Let  A  :  B  :  :  B  :  C  ;  then  will  A  :  C  :  :  A^  :  B^ 

For,  from  the  given  proportion,  by  Prop.  III.,  we  have 

A  X  C  =  B^ 

Multiplying  each  side  of  this  equation  by  A  gives 
A^  X  C  =  A  X  B^ 

Hence,  by  Prop.  II., 

A  :  C  :  :  A'*  :  B^ 

5* 


54  ELEMENTS    OF    GEOMETRY. 


Proposition  XV.  —  Theorem. 

151.  If  four  magnitudes  are  proportionals^  their  like 
powers  and  roots  will  also  be  proportional. 

Let  A  :  B  :  :  C  :  D  ;  then  will 

A"  :  B"  :  :  C"  :  D%      and      A^^  :  B^  :  :  C*  :  DK 

For,  from  the  given  proportion,  we  have 

A  _  C 

B  ~  D* 

Raising  both  members  of  this  equation  to  the  wtli  power, 
we  have 

^  _  51 

and  extracting  the  n\h  root  of  each  member,  we  have 

A*  _  C^ 

B^  ~~  D^' 
Hence,  by  Prop.  II.,  the  last  two  equations  give 

A^'  :  B'^  :  :  C'  :  D", 
and 

A^  :  B»  :  :  C»  :  D». 


BOOK    III 


THE   CIRCLE,  AND  THE  MEASURE  OF   ANGLES. 


DEFINITIONS. 

152.  A  CIRCLE  is  a  plane  fig^ure 
bounded  by  a  curved  line,  all  tlie 
points  of  wliicli  are  equally  distant 
from  a  point  witbin  called  tlie  centre; 
as  tbe  figure  A  D  B  E. 


153.  Tbe  CIRCUMFERENCE  or  PERIPHERY  of  a  circle  is  its 
entire  bounding  line  ;  or  it  is  a  curved  line,  all  points  of 
wbicb  are  equally  distant  from  a  point  witbin  called  tbe 
centre. 

154.  A  RADIUS  of  a  circle  is  any  straigbt  line  drawn 
from  tbe  centre  to  tbe  circumference  ;  as  tbe  line  C  A, 
CD,  or  CB. 

155.  A  DIAMETER  of  a  circlc  is  any  straigbt  line  drawn 
tbrougb  tbe  centre,  and  terminating  in  botb  directions  in 
tbe  circumference  ;  as  tbe  line  A  B. 

All  tbe  radii  of  a  circle  are  equal ;  all  tbe  diameters  are 
also  equal,  and  eacb  is  double  tbe  radius. 

156.  An  ARC  of  a  circle  is  any  part 
of  tbe  circumference  ;  as  tbe  part 
AD,  AE,  or  EGF. 

157.  Tbe  CHORD  of  an  arc  is  tbe 
straigbt  line  joining  its  extremities  ; 
tbus  E  F  is  tbe  cbord  of  tbe  arc 
EGF. 


66 


ELEMENTS   OF   GEOMETRY. 


158.  The  SEGMENT  of  a  circle  is 
the   part   of  a   circle   iiichided    be- 
tween an  arc  and  its  chord  ;  as  the 
surface    included    between   the   arc  ^ 
EGF  and  the  chord  EF.  E 

159.  The  SECTOR  of  a  circle  is  the 
part  of  a  circle  included  between  an  ^ 

arc,  and  the  two  radii  drawn  to  the  eHtremities  of  the 
arc  ;  as  the  surface  included  between  the  arc  A  D,  and 
the  two  radii  CA,  CD. 


160.  A  SECANT  to  a  circle  is  a 
straight  line  which  meets  the  cir- 
cumference in  two  points,  and  lies 
partly  within  and  partly  without 
the  circle  ;  as  the  line  A  B. 


161.  A  TANGENT  to  a  circle  is  a  straight  line  which, 
how  far  so  ever  produced,  meets  the  circumference  in  but 
one  point;  as  the  line  CD.  The  point  of  meeting  is 
called  the  point  of  contact  ;  as  the  point  M. 

162.  Two  circumferences  touch 
each  other,  when  they  have  a  point  /-— ^\  /  ^ — -^ 
of  contact  without  cutting  one  an- 
other ;  thus  two  circumferences 
touch  each  other  at  the  point  A, 
and  two  at  the  point  B. 

163.  A  STRAIGHT  LINE  is  IN- 
SCRIBED in  a  circle  when  its  ex- 
tremities are  in  the  circumference; 
as  the  line  A  B,  or  B  C. 

164.  An  INSCRIBED  ANGLE  is  onc  which  has  its  vertex  in 
the  circumference,  and  is  formed  by  two  chords ;  as  the 
angle  ABC. 


BOOK   III. 


57 


165.   An  INSCRIBED  POLYGON  is  oiie 

which  has  the  vertices  of  all  its  angles 
in  the  circumference  of  the  circle  ; 
as  the  triangle  ABC. 


166.  The  circle  is  then  said  to  be  circumscribed  about 
the  polygon. 


167.    A  POLYGON   is  CIRCUMSCRIBED 

about  a  circle  when  all  its  sides  are 
tangents  to  the  circumference ;  as 
the  polygon  ABCDEF. 


168.  The  circle  is  then  said  to  be  inscribed  in  the 
polygon. 

Proposition  I. — Theorem. 

169.  Every  diameter  divides  the  circle  and  its  circum- 
ference each  into  two  equal  parts. 

Let  A  E  B  F  be  a  circle,  and  A  B  F 

a    diameter ;    then    the    two    parts 
A  E  B,  A  F  B  are  equal. 

For,  if  the  figure  A  E  B  be  applied 
to  AF  B,  tlieir  common  base  A  B  re- 
taining its  position,  the  curve  line 
A  E  B  must  fall  exactly  on  the  curve 
lino  AFB;  otherwise  there  would  be 
points  in  the  one  or  the  other  unequally  distant  from  the 
centre,  which  is  contrary  to  the  definition  of  the  circle 
(Art.  152).  Hence  a  diameter  divides  the  circle  and  its 
circumference  into  two  equal  parts. 

170.  Cor.  1.     Conversely^  a  straight  line  dividing  the 
circle  into  two  equal  parts  is  a  diameter. 


58  ELEMENTS    OF   GEOMETRY. 

For,  let  the  line  AB  divide  the 

■p 

circle  A E  B  C  F  into  two  equal  parts ;  ^ — ^^^ 

then,  if  the  centre  is  not  in  A  B,  let  /^               N. 

A  C  be  drawn  through  it,  which  is  /                   \  ^ 

therefore    a    diameter,    and    conse-  Ar"" ^'b 

quently  divides  the  circle  into  two  \                        j 

equal  parts ;  hence  the  surface  AFC  \^               / 

is  equal  to  the  surface  A  F  C  B,  a  part  ^^~E^^^ 
to  the  whole,  which  is  impossible. 

171.  Cor.  2.  The  arc  of  a  circle,  whose  chord  is  a 
diameter,  is  a  semi-circumference,  and  the  included  seg- 
ment is  a  semicircle. 

Pkoposition   II.  —  Theorem. 

172.  A  straight  line  cannot  meet  the  circumference  of 
a  circle  in  more  than  two  points. 

For,  if  a  straight  Ihie  could  meet 
the  circumference  ABD,  in  three 
points,  A,  B,  D,  join  each  of  these 
points  with  the  centre,  C  ;  then, 
since  the  straight  lines  C  A,  C  B, 
C  D  are  radii,  they  are  equal  (Art. 
155)  ;  hence,  three  equal  straight 
lines  can  be  drawn  from  the  same  point  to  the  same 
straight  line,  which  is  impossible  (Prop.  XIV.  Cor.  2, 
Bk.  L). 

Proposition  III.  —  Theorem. 

173.  In  the  same  circle^  or  in  equal  circles,  equal  arcs 
are  subtended  by  equal  chords;  and,  conversely,  equal 
chords  subtend  equal  arcs. 

Let  A  D  B  and  E  G  F  be  two  equal  circles,  and  let  the 
arc  A  D  be  equal  to  E  C ;  then  will  the  chord  A  D  be 
equal  to  the  chord  E  G. 


'    BOOK   III.  69 

For,  since  the 
diameters  A  B, 
EF    are  equal, 

the     semicircle     [  r  i^  -^ 

A  D  B  may  be 
applied  to  the 
semicircle  E  G  F ; 
and  tlie  curve  line  A  D  B  will  coincide  with  the  curve 
line  EGF  (Prop.  I.).  But,  by  hypothesis,  the  arc  AD 
is  equal  to  the  arc  E  G ;  hence  the  point  D  will  fall  on  G ; 
hence  the  chord  A  D  is  equal  to  the  chord  E  G  (Art. 
34,  Ax.  11). 

Conversely^  if  the  chord  A  D  is  equal  to  the  chord  E  G, 
the  arcs  A  D,  E  G  will  be  equal. 

For,  if  the  radii  CD,  0 G  are  drawn,  the'  triangles 
A  C  D,  E  0  G,  having  the  three  sides  of  the  one  equal  to 
the  three  sides  of  the  other,  each  to  each,  are  themselves 
equal  (Prop.  XVIII.  Bk.  I.)  ;  therefore  the  angle  A  C  D 
is  equal  to  the  angle  E  0  G  (Prop.  XVIII.  Sch.,  Bk.  I.). 

If  now  the  semicircle  A  D  B  be  applied  to  its  equal 
EGF,  with  the  radius  A C  on  its  equal  E 0,  since  the 
angles  A  C  D,  E  0  G  are  equal,  the  radius  C  D  will  fall  on 
OG,  and  the  point  D  on  G.  Therefore  the  arcs  AD 
and  EG  coincide  with  each  other ;  hence  they  must  be 
equal  (Art.  34,  Ax.  14). 

Proposition  IV.  —  Theorem. 

174.  In  the  same  circle^  or  in  equal  circles,  a  greater 
arc  is  subtended  by  a  greater  chord;  and,  conversely,  the 
greater  chord  subtends  the  greater  arc. 

In  the  circle  of  which  C  is  the  centre,  let  the  arc  A  B 
1)0  greater  than  the  arc  A  D  ;  then  will  the  chord  A  B  be 
greater  than  the  chord  AD. 

Draw  the  radii  CA,  CD,  and  C  B.     Tlie  two  sides  AC^ 


GO 


ELEMENTS    OF   GEOlvfETRY. 


C  B  in  the  triangle  A  C  B  are  equal 
to  the  two  A  C,  C  D  in  the  triangle 
A  CD,  and  the  angle  ACB  is  greater 
than  the  angle  A  C  D  ;  therefore  the 
third  side  A  B  is  gi-eater  than  the 
third  side  A  D  (Prop.  XVI.  Bk.  I.)  ; 
hence  the  chord  which  subtends  the 
greater  arc  is  the  greater. 

Conversely^  if  the  chord  A  B  be  greater  than  the  chord 
AD,  the  arc  A  B  will  be  greater  than  the  arc  AD. 

For  the  triangles  ACB,  ACD  have  two  sides,  AC,  CB, 
in  the  one,  equal  to  two  sides,  AC,  CD,  in  the  other,  while 
the  side  A  B  is  greater  tlian  the  side  AD  ;  therefore  the 
angle  AQB  is  greater  than  the  angle  ACD  (Prop.  XVII. 
Bk.  I.)  ;  hence  the  arc  A B  is  greater  than  the  arc  AD. 

175.  Scholium.  Tlie  arcs  here  treated  of  are  each  less 
than  the  senii-circumfercnce.  If  they  were  greater,  the 
contrary  would  be  true  ;  in  which  case,  as  the  arcs  in- 
creased, the  chords  would  diminish,  and  conversely. 


Proposition  V.  —  Theorem. 

176.  In  the  same  circle^  or  in  equal  circles^  radii  ivhich 
make  equal  ang-Ies  at  the  centre  intercept  equal  arcs  on 
the  circumference ;  and,  conversely,  if  the  intercepted 
arcs  are  equal,  the  ang-les  made  by  the  radii  are  also 
equal. 

Let  ACB  and 
I)  C  E  be  equal  angles 
made  by  radii  at  the 
centre  of  equal  cir- 
cles ;  then  will  the  \ 
intercepted  arcs  A  B  /, 
and  DE  be  also  equal. 

First.  Since  the  angles  A  0  B,  D  C  E  are  equal,  the 
one  may  bo  applied  to  the  otlicr;  and  since  their  sides, 


BOOK   III. 


61 


being  radii  of  equal  circles,  are  equal,  the  point  A  will 
coincide  with  D,  and  the  point  B  with  E.  Therefore  the 
arc  A  B  must  also  coincide  with  the  arc  D  E,  or  there 
would  be  points  in  the  one  or  tlie  other  unequally  distant 
from  the  centre,  which  is  impossible  ;  hence  tiie  arc  A  B 
is  equal  to  the  arc  D  E. 

Second.  If  the  arcs  A  B  and  D  E  are  equal,  the  angles 
A  C  B  and  D  C  E  will  be  equal. 

For,  if  these  angles  are  not  equal,  let  ACB  be  tha 
greater,  and  let  A  C  F  be  taken  equal  to  D  C  E.  From 
what  has  been  shown,  we  shall  have  the  arc  AF  equal  to 
the  arc  D  E.  But,  by  hypothesis,  A  B  is  equal  to  D  E  ; 
hence  A  F  must  be  equal  to  A  B,  the  part  to  the  whole, 
which  is  impossible  ;  hence  the  angle  A  C  B  is  equal  to 


the  angle  D  C  E. 


Proposition  YI.  —  Theorem. 


177.    The  radius  which  is  perpendicular  to  a  chord  bi- 
sects the  chord,  and  also  the  arc  subtended  by  the  chord. 

Let  the  radius  C  E  be  perpendicu- 
lar to  the  chord  A  B  ;  then  will  C  E 
bisect  the  chord  at  D,  and  the  arc 
AB  at  E. 

Draw  the  radii  C  A  and  C  B. 
Then  C  A  and  C  B,  with  respect  to 
tlie  perpendicular  C  E,  are  equal 
ol)lique  lines  drawn  to  the  chord  AB ; 
therefore  their  extremities  are  at  equal  distances  from  the 
perpendicular  (Prop.  XIV.  Bk.  I.)  ;  hence  A  I)  and  I)  B 
are  equal. 

Again,  since  the  triangle  ACB  has  the  sides  A  C  and 
CB  equal,  it  is  isosceles;  and  the  line  CE  bisects  the 
base  A  B  at  right  angles  ;  therefore  C  E  bisects  also  the 
angle  ACB  (Prop.  VII.  Cor.  2,  Bk.  I.).  Since  the  an- 
gles A  C  D,  D  C  B  are  equal,  the  arcs  A  E,  E  B  are  equal 
6 


62  ELEMENTS   OP   GEOMETRY. 

(Prop.  Y.)  ;  hence  the  radius  C  E,  which  is  perpendicular 
to  the  chord  A  B,  bisects  the  arc  A  B  subtended  by  the 
chord. 

178.  Cor.  1.  Any  straight  line  which  joins  the  centre 
of  the  circle  and  the  middle  of  the  chord,  or  the  middle 
of  the  arc,  must  be  perpendicular  to  the  chord. 

For  the  perpendicular  from  the  centre  C  passes  through 
tlie  middle,  D,  of  the  chord,  and  the  middle,  E,  of  the  arc 
subtended  by  the  chord.  Now,  any  two  of  these  three 
points  in  the  straight  line  C  E  are  sufficient  to  determine 
its  position. 

179.  Cor.  2.  A  perpendicular  at  the  middle  of  a  chord 
passes  through  the  centre  of  tlie  circle,  and  through  the 
middle  of  the  arc  subtended  by  the  chord,  bisecting  at  the 
centre  the  angle  which  the  arc  subtends. 

Proposition  VII.  —  Theorem. 

180.  Through  three  given  points,  not  in  the  same 
straight  line,  one  circumference  can  he  made  to  pass, 
and  but  one. 

Let  A,  B,  and    C   be   any   three  ^ .^^ 

points  not  in  the  same  straight  line  ;  /^  ^v 

one  circumference  can  be  made  to         /  \  /  \ 

pass  through  them,  and  but  one.  \^\.  / 

_  Join   A  B    and   B  C ;    and    bisect        \        V      J^      I 
these  straight  lines  by  the  perpendic-         \       \    /       / 
ulars  D  E  and  F  E.    Join  J)  F  ;  then,  ^^^^-^ 

the  angles  B  D  E,  B  F  E,  being  each 
a  right  angle,  are  together  equal  to  two  right  angles ; 
therefore  the  angles  E  D  F,  E  F  I)  are  togetlier  loss  than 
two  right  angles;  hence  D  E,  FE,  produced,  must  meet 
in  some  point  E  (Prop.  XXIII.  Bk.  I.). 

Now,  since  the  point  E  lies  in  the  perpendicular  D  E,  it 
is  equally  distant  from  the  two  points  A  and  B  (Prop. 
XY.  Bk.  I.)  ;  and  since  the  same  point  E  lies  in  the  per- 


BOOK  III.  63 

pendicular  FE,  it  is  also  equally  distant  from  the  two 
points  B  and  C  ;  therefore  the  three  distances,  E  A,  E  B, 
E  C,  are  equal ;  hence  a  circumference  can  be  described 
from  the  centre  E  passing  through  the  three  points  A,  B,  C. 
Again,  the  centre,  lying  in  the  perpendicular  D  E  bi- 
secting the  chord  AB,  and  at  the  same  time  in  the  per- 
pendicular F  E  bisecting  the  chord  B  C  (Prop.  VI.  Cor. 
2),  must  be  at  the  point  of  their  meeting,  E.  There- 
fore, since  there  can  be  but  one  centre,  but  one  circum- 
ference can  be  made  to  pass  through  three  given  points. 

181.  Cor.  Two  circumferences  can  intersect  in  only 
two  points ;  for,  if  they  have  three  points  in  common,  they 
must  have  the  same  centre,  and  must  coincide. 

Proposition  VIII.  —  Theorem. 

182.  Equal  chords  are  equally  distant  from  the  centre  ; 
and  J  conversely^  chords  ivhich  are  equally  distant  from  the 
centre  are  equal. 

Let  A  B  and  D  E  bo  equal  chords, 
and  C  the  centre  of  the  circle  ;  and 
draw  C  F  perpendicular  to  A  B,  and 
C  G  perpendicular  to  D  E ;  then 
these  perpendiculars,  which  measure 
the  distance  of  the  chords  from  the 
centre,  are  equal. 

Join  CA  and  CD.  Then,  in  the  right-angled  triangle 
C  AF,  C  D  G,  the  hypothenuses  C  A,  C  D  are  equal ;  and 
the  side  A  F,  the  half  of  A  B,  is  equal  to  the  side  D  G,  the 
half  of  D  E  ;  therefore  the  triangles  are  equal,  and  C  F  is 
equal  to  C  G  (Prop.  XIX.  Bk.  I.)  ;  hence  the  two  equal 
chords  A  B,  D  E  are  equally  distant  from  the  centre. 

Conversely^  if  the  distances  C  F  and  C  G  are  equal,  the 
chords  A  B  and  D  E  are  equal. 

For,  in  the  right-angled  triangles  A  C  F,  D  C  G,  the 
hypothenuses  C  A,  C  D  are  equal ;  and  the  side  C  F  is 


64 


ELEMENTS   OF   GEOMETRY. 


equal  to  the  side  C  G  ;  therefore  the  triangles  are  equal, 
and  AF  is  equal  to  D  Gr ;  hence  AB,  the  dovible  of  AF, 
is  equal  to  D  E,  the  double  of  D  G  (Art.  34,  Ax.  6). 


Proposition  IX.  —  Theorem. 

183.    Of  tivo  unequal  chords ,  the  less   is   the  farther 
from  the  centre. 

Of  the  two  chords  D  E  and  A II, 
let  A  H  be  the  greater  ;  then  will 
D  E  be  the  farther  from  the  cen- 
tre C. 

Since  the  chord  AH  is  greater 
than  the  chord  D  E,  the  arc  A  H  is 
greater  than  the  arc  D  E  (Prop. 
IV.).  Cut  off  from  the  arc  A  H  a  part,  A  B,  equal  D  E  ; 
draw  CF  perpendicular  to  this  chord,  CI  perpendicular 
to  AH,  and  C  G  perpendicular  to  D  E.  C  F  is  greater 
than  C  0  (Art.  34,  Ax.  8),  and  C  0  than  C  I  (Prop.  XIV. 
Bk.  I.)  ;  therefore  C  F  is  greater  than  C  I.  But  C  F  is 
equal  to  C  G,  since  the  chords  A  B,  D  E  are  equal  (Prop. 
VIII.)  ;  therefore,  C  G  is  greater  than  C  I ;  hence,  of  two 
unequal  chords,  the  less  is  the  farther  from  the  centre. 


—  D 


Proposition  X.  —  Theorem. 

184.  A  straig-ht  line  perpendicular  to  a  radius  at  its 
termination  in  the  circu/mference,  is  a  tang-ent  to  the  circle. 

Let  the  straight  line  B  D  be  per- 
pendicular to  the  radius  C  A  at  its 
termination  A ;  then  will  it  be  ia 
tangent  to  the  circle. 

Draw  from  the  centre  C  to  BD 
any  other  straight  line,  as  C  E. 
Then,  since  C  A  is  perpendicular  to 
B  D,  it  is  shorter  than  the  oblique 


BOOK  in.  65 

line  C  E  (Prop.  XIY.  Bk.  I.)  ;  hence  the  point  E  is  with- 
out the  circle.  The  same  may  be  shown  of  any  other 
point  in  the  line  BD,  except  the  point  A ;  tlierefore  BD 
meets  the  circumference  at  A,  and,  being  produced,  does 
not  cut  it ;  hence  B  D  is  a  tangent  (Art.  161). 

Proposition  XI.  —  Theorem. 

185.  If  a  line  is  A  tangent  to  a  circumference^  the  ra- 
dius drawn  to  the  point  of  contact  with  it  is  perpendicular 
to  the  tangent. 

Let  B  D  be  a  tangent  to  the  cir- 
cumference, at  the  point  A ;  then 
will  the  radius  C  A  be  perpendicu- 
lar to  BD. 

For  every  point  in  BD,  except  A, 
being  without  the  circumference 
(Prop.  X.),  any  line  CE  drawn 
from  the  centre  C  to  B  D,  at  any 
point  other  than  A,  must  terminate  at  E,  without  the  cir- 
cumference ;  therefore  the  radius  C  A  is  the  shortest  line 
that  can  be  drawn  from  the  centre  to  B  D  ;  hence  C  A  is 
perpendicular  to  the  tangent  B  D  (Prop.  XIY.  Cor.  1, 
Bk.  I.). 

186.  Cor.  Only  one  tangent  can  be  drawn  through 
the  same  point  in  a  circumference  ;  for  two  lines  cannot 
both  be  perpendicular  to  a  radius  at  the  same  point. 

Proposition  XII.  —  Theorem. 

187.  Tivo  parallel  straight  lines  intercept  equal  arcs 
of  the  circumference. 

First.  When  the  two  parallels  are  secants,  as  AB,  DE. 

Draw  the  radius  C  H  perpendicular  to  A  B  ;  and  it  will 

also  be  perpendicular  to  D  E  (Prop.  XXII.  Cor.,  Bk.  I.)  ; 

6* 


G6 


ELEMENTS   OF   GEOMETRY. 


therefore  the  point  H  will  be  at  the 
same  time  the  middle  of  the  arc 
A  H  B  and  of  the  arc  D  H  E  (Prop. 
YI.)  ;  therefore,  the  arc  AH  is  equal 
to  the  arc  H  B,  and  the  arc  D  H  is 
equal  to  the  arc  H  E  ;  hence  AH 
diminished  by  D  H  is  equal  to  H  B 
diminished  by  H  E  ;  that  is,  the  in- 
tercepted arcs  AD,  BE  are  equal. 

Second.  When  of  the  two  parallels,  one,  as  AB,  is  a 
secant,  and  the  other,  as  D  E,  is  a  tangent. 

Draw  the  radius  C  H  to  the  point 
of  contact  H.  This  radius  will  be  D 
perpendicular  to  the  tangent  D  E 
(Prop.  X.),  and  also  to  its  parallel 
AB  (Prop.  XXn.  Cor.,  Bk.  I.). 
But,  since  C  H  is  perpendicular  to 
the  chord  AB,  the  point  H  is  the 
middle  of  the  arc  A  H  B  ;  hence  the 
arcs  AH,  HB,  included  between 
the  parallels  AB,  D  E,  are  equal. 

Third.  When  the  two  parallels  are  tangents,  as  D  E, 
IL. 

Draw  the  secant  AB  parallel  to  either  of  the  tangents, 
and  it  will  be  parallel  to  the  other  (Prop.  XXI Y.  Bk.  I.)  ; 
then,  from  what  has  been  just  shown,  the  arc  AH  is  equal 
to  the  arc  H  B,  and  also  the  arc  A  G  is  equal  to  the  arc 
G  B  ;  hence  the  whole  arc  HAG  is  equal  to  the  whole 
arc  H  B  G. 

It  is  further  evident,  since  the  two  arcs  HAG,  H  B  G 
are  equal,  and  together  make  up  the  whole  circumference, 
that  each  of  them  is  a  semi-circumference. 

188.  Cor.  Two  parallel  tangents  meet  the  circumfer- 
ence at  the  extremities  of  the  same  diameter. 


BOOK   III. 


67 


Proposition  XIII.  —  Theorem. 

189.  If  tivo  circumferences  touch  each  other  externally 
or  internally^  their  centres  and  the  point  of  contact  are 
in  the  same  straight  line. 

Let  the  two  circumferences, 
whose  centres  are  C  and  D, 
touch  each  other  externally  in 
the  point  A ;  the  points  C,  B, 
and  A  will  be  all  in  the  same 
straight  line. 

Draw  from  the  point  of  con- 
tact A  the  common  tangent  A  B.  Then  the  radius  C  A  of 
tlie  one  circle,  and  the  radius  D  A  of  the  other,  are  each 
perpendicular  to  A  B  (Prop.  XI.)  ;  but  there  can  be  but 
one  straight  line  drawn  through  the  point  A  perpendicular 
to  AB  (Prop.  XIII.  Bk.  I.)  ;  therefore  the  points  C,  D, 
and  A  are  in  one  perpendicular ;  hence  they  are  in  one 
and  the  same  straight  line. 

Also,  let  the  two  circumferences 
touch  each  other  internally  in  A ; 
tlion  their  centres,  C  and  D,  and  the 
point  of  contact,  A,  will  be  in  the 
same  straight  line. 

Draw  the  common  tangent  AB. 
Then  a  straight  line  perpendicular 
to  A  B,  at  the  point  A,  on  being  suf- 
ficiently produced,  must  pass  through  the  two  centres  C 
and  D  (Prop.  XI.)  ;  but  from  the  same  point  there  can 
be  but  one  perpendicular  ;  therefore  the  points  C,  D,  and 
A  are  in  that  perpendicular ;  hence  they  are  in  the  same 
straight  line. 

190.  Cor.  1.  When  two  circumferences  touch  each 
other  externally,  the  distance  between  their  centres  is 
equal  to  the  sum  of  their  radii. 


68 


ELEMENTS   OF   GEOMETRY. 


191.  Cor.  2.  And  when  two  circumferences  touch  each 
other  internally,  the  distance  between  their  centres  is 
equal  to  the  difference  of  their  radii. 

Proposition  XIY. — Theorem. 

192.  ^  two  circumferences  cut  each  other,  the  straight 
line  passing  through  their  centres  will  bisect  at  right  an- 
gles the  chord  which  joins  the  points  of  intersection. 

Let  two  circumferences  cut  each  other  at  the  points 
A  and  B;   then  the  straight  line  passing  through   the 


centres  C  and  D  will  bisect  at  right  angles  the  chord  A  B 
common  to  the  two  circles. 

For,  if  a  perpendicular  be  erected  at  the  middle  of  this 
chord,  it  will  pass  through  each  of  the  two  centres  C  and 
D  (Prop.  VI.  Cor.  1).  But  no  more  than  one  straight 
line  can  be  drawn  through  two  points  ;  hence  the  straight 
line  C  D,  passing  through  the  centres,  must  bisect  at  right 
angles  the  common  chord  A  B. 

193.  Cor.  The  straight  line  joining  the  points  of  inter- 
section of  two  circumferences  is  perpendicular  to  the 
straight  line  which  passes  through  their  centres. 


Proposition  XY.  —  Theorem. 

194.  If  two  circumferences  cut  each  other,  the  distance 
betiveen  their  centres  ivill  be  less  than  the  sum  of  their 
radii,  and  greater  than  their  difference. 


BOOK    III. 


69 


Let  two  circumferences 
whose  centres  are  C  and 
D  cut  each  other  in  the 
point  A,  and  draw  the 
radii  C  A  and  D  A.  Then, 
in  order  that  the  intersec- 
tion may  take  place,  the 
triangle  CAD  must  be  possible.  And  in  this  triangle  the 
side  CD  must  be  less  tlian  the  sum  of  AC  and  AD 
(Prop.  IX.  Bk.  I.)  ;  also  C  D  must  be  greater  than  the 
difference  between  DA  and  C A  (Prop.  IX.  Cor.,  Bk.  I.). 


Proposition  XYI.  —  Theorem. 

195.  In  the  same  circle^  or  in  equal  circles^  if  two  an- 
gles at  the  centre  are  to  each  other  as  tiuo  ivhole  numbers, 
the  intercepted  arcs  will  be  to  each  other  as  the  same 
numbers. 

Let  us  suppose,  for  example,  that  the  angles  ACB, 
D  C  E,  at  the  centre  of  equal  circles,  are  to  each  other  as 
7  to  4  ;  or,  which  amounts  to  the  same  thing,  that  the 
angle  M,  which  will  serve  as  a  common  measure,  is  con- 


tained seven  times  in  the  angle  ACB,  and  four  times  in 
the  angle  D  C  E.  The  seven  partial  angles  A  C  m,  mCn, 
nCp,  &G.  into  which  ACB  is  divided,  being  each  equal 
to  any  of  the  four  partial  angles  into  whicli  D  C  E  is 
divided,  each  of  the  partial  arcs  A  m,  m  n,  np,  <fec.  will 


70  ELEMENTS  OF  GEOMETRY. 

be  also  equal  to  each  of  the  partial  arcs  Dx,  xy^  <fec. 
(Prop.  V.)  ;  therefore  the  whole  arc  A  B  will  be  to  the 
whole  arc  D  E  as  7  to  4.  But  the  same  reasoning  would 
apply,  if  in  place  of  7  and  4  any  numbers  whatever  were 
employed  ;  hence,  if  the  ratio  of  the  angles  A  C  B,  D  C  E 
can  be  expressed  in  whole  numbers,  the  arcs  A  B,  D  E 
will  be  to  each  other  as  the  angles  ACB,  DCE. 

196.  Cor.  Conversely^  if  the  ai'cs  A  B,  D  E  are  to  each 
other  as  two  whole  numbers,  the  angles  A  C  B,  D  C  E  will 
be  to  each  other  as  the  same  whole  numbers,  and  we  shall 
have  ACB:DCE::AB:DE.  For,  the  partial  arcs 
A?M,  mw,  <fec.  and  Da:,  .^•y,  &c.  being  equal,  the  partial 
angles  A  C  m^  m  C  w,  &c,  and  D  C  x,  x  C «/,  <fec.  will  also 
be  equal. 

Proposition  XVII.  —  Theorem. 

197.  In  the  same  circle,  or  in  equal  circles,  any  two 
ang-les  at  the  centre  are  to  each  other  as  the  arcs  inter- 
cepted hetiveen  their  sides. 

Let  ACB  be 
the  greater,  and 
A  C  D  the  less 
angle  ;  then  will 
the  angle  ACB 
be  to  the  angle 
A  C  D  as  the  arc 
A  B  is  to  the  arc 
AD. 

Conceive  the  less  angle  to  be  placed  on  the  greater ; 
then,  if  the  proposition  be  not  true,  the  angle  ACB  will 
be  to  the  angle  A  C  D  as  the  arc  A  B  is  to  an  arc  greater 
or  less  than  AD.  Suppose  this  arc  to  be  greater,  and  let 
it  be  represented  by  A  0  ;  we  shall  have  the  angle  ACB: 
angle  A  C  D  :  :  arc  A  B  :  arc  A  0.  Conceive,  now,  the  arc 
A  B  to  be  divided  into  equal  parts,  each  of  which  is  less 


BOOK   III.  71 

than  D  0  ;  there  will  be  at  least  one  point  of  division  be- 
tween D  and  0  ;  let  I  be  that  point ;  and  join  C  I.  The 
arcs  A  B,  A I  will  be  to  each  other  as  two  whole  numbers, 
and,  by  tlie  preceding  proposition,  we  shall  have  tlie  an- 
gle A  C  B  :  angle  A  C  I  :  :  arc  A  B  :  arc  A  I.  Comparing 
these  two  proportions  with  each  other,  and  observing  that 
tlie  antecedents  are  the  same,  we  infer  that  the  conse- 
quents are  proportional  (Prop.  X.  Cor.  2,  Bk.  II.)  ;  hence 
the  angle  ACD  :  angle  ACl  :  :  arc  AO  :  arc  A  I.  But 
tlic  arc  A  0  is  greater  than  the  arc  A I ;  therefore,  if  this 
proportion  is  true,  the  angle  ACD  must  be  greater  than 
the  angle  A  C  I.  But  it  is  less  ;  hence  the  angle  A  C  B 
cannot  be  to  the  angle  A  C  D  as  the  arc  A  B  is  to  an  arc 
greater  than  AD. 

By  a  process  of  reasoning  entirely  siniihir,  it  may  be 
shown  that  the  fourth  term  of  the  proportion  cannot  be 
less  than  A  D  ;  therefore  it  must  be  A  D  ;  hence  we  have, 

Angle  A  C  B  :  angle  A  C  D  :  :  arc  A  B  :  arc  A  D. 

198.  Scholium  1.  Since  the  angle  at  the  centre  of  a 
circle,  and  the  arc  intercepted  by  its  sides,  have  sucli  a 
connection,  that,  if  the  one  be  increased  or  dimhiished  in 
any  ratio,  the  other  will  be  increased  or  diminished  in  the 
same  ratio,  we  are  authorized  to  take  the  one  of  tliese 
magnitudes  as  the  measure  of  the  other.  Henceforth  we 
shall  assume  the  arc  AB  as  the  measure  of  the  angle 
A  C  1^  It  is  to  be  observed,  in  the  comparison  of  angles 
with  each  other,  that  the  arcs  which  serve  to  measure 
them  must  be  described  with  equal  radii. 

199.  Scholium  2.  Sectors  taken  in  the  same  circle,  or 
in  equal  circles,  arc  to  each  other  as  their  arcs  ;  for  sec- 
tors are  equal  wlicn  tlieir  angles  are  so,  and  therefore  are 
in  all  respects  proportional  to  their  angles. 


72 


ELEMENTS    OF   GEOMETRY. 


Proposition  XVIII.  —  Theorem. 

200.  An  inscribed  angle  is  measured  by  half  the  arc 
included  between  its  sides. 

Let  B  A  D  be  an  inscribed  angle, 
whose  sides  include  the  arc  B  D  ; 
then  the  angle  BAD  is  measured 
by  half  of  the  arc  B  D. 

First.  Suppose  the  centre  of  the 
circle  C  to  lie  within  the  angle 
BAD.  Draw  the  diameter  AE, 
and  the  radii  CB,  CD. 

The  angle  B  C  E,  being  exterior  to  the  triangle  ABC, 
is  equal  to  the  sum  of  the  two  interior  angles  CAB,  ABC 
(Prop.  XXVII.  Bk.  I.).  But  the  triangle  BAC  being 
isosceles,  the  angle  C  A  B  is  equal  to  A  B  C  ;  hence,  the 
angle  B  C  E  is  double  BAC.  Since  B  C  E  lies  at  the 
centre,  it  is  measured  by  the  arc  B  E  (Prop.  XVII.  Sch. 
1)  ;  hence  BAC  will  be  measured  by  half  of  B  E.  For  a 
like  reason,  the  angle  CAD  will  be  measured  by  the  half 
of  ED;  hence  BAC  and  CAD  together,  or  BAD,  will 
be  measured  by  the  half  of  B  E  and  E  D,  or  half  B  D. 

Second.  Suppose  that  the  centre  ^ 

C  lies  without  the  angle  BAD. 
Then,  drawing  the  diameter  AE, 
the  angle  B  A  E  will  be  measured 
by  the  half  of  B  E  ;  and  the  angle 
D  A  E  is  measured  by  the  half  of 
D  E  ;  hence,  their  difference,  BAD, 
will  be  measured  by  the  half  of  BE 
minus  the  half  of  ED,  or  by  the 
half  of  BD. 

Hence  every  inscribed  angle  is  measured  by  the  half  of 
the  arc  included  between  its  sides. 


BOOK   III. 


73 


201.  Cor,  1.  All  the  angles, 
BAG,  BDC,  inscribed  in  the 
same  segment,  are  equal ;  because 
they  are  all  measured  by  the  half 
of  the  same  arc,  BOG. 


202.  Cor.  2.  Every  angle,  BAD, 
inscribed  in  a  semicircle,  is  a  right 
angle  ;  because  it  is  measured  by 
half  the  semi-circumference,  BOD ; 
that  is,  by  the  fourth  part  of  the 
whole  circumference. 

203.  Cor.  3.  Every  angle,  BAG, 
inscribed  in  a  segment  greater  than 
a  semicircle,  is  an  acute  angle ;  for 
it  is  measured  by  the  half  of  the 
arc  BOG, less  than  a  semi-circum- 
ference. 

And  every  angle,  BOG,  inscribed 
in  a  segment  less  than  a  semicircle, 
is  an  obtuse  angle  ;  for  it  is  meas- 
ured by  half  of  the  arc  BAG, 
greater  than  a  semi-circumference. 

204.  Cor.  4.  The  opposite  an- 
gles, A  and  D,  of  an  inscribed 
quadrilateral,  ABD  G,  are  together 
equal  to  two  right  angles  ;  for  the 
angle  BAG  is  measured  by  half 
the  arc  BDG,  and  the  angle  BDG 
is  measured  by  half  the  arc  BAG; 
hence  the  two  angles  BAG,  BDG,  taken  together,  are 
measured  by  half  the  circumference  ;  hence  their  sum 
is  equal  to  two  right  angles. 

7 


74 


ELEMENTS   OF   GEOMETRY. 


Proposition  XIX.  —  Theorem. 

205.  The  angle  formed  by  the  inter sectio7i  of  two  chords 
is  measured  by  half  the  sum  of  the  tivo  intercepted  arcs. 

Lot  the  two  chords  A  B,  C  D  mter- 
sect  each  other  at  the  point  E ;  then 
will  the  angle  DEB,  or  its  equal, 
A  E  C,  be  measured  by  half  the  sum 
of  the  two  arcs  D  B  and  A  C. 

Draw  A  F  parallel  to  D  C  ;  then 
will  the  arc  FD  be  equal  to  the 
arc  AC  (Prop.  XII.),  and  the  an- 
gle FAB  equal  to  the  angle  DEB  (Prop.  XXII.  Bk.  I.). 
But  the  angle  F  A  B  is  measured  by  half  the  arc  F  D  B 
(Prop.  XVIII.) ;  that  is,  by  half  the  arc  D  B,  plus  half 
the  arc  FD.  Hence,  since  FD  is  equal  to  A  C,  the  angle 
DEB,  or  its  equal  angle  A  EC,  is  measured  by  half  the 
sum  of  the  intercepted  arcs  D  B  and  A  C 


Proposition  XX.  —  Theorem. 

206.   The  angle  formed  by  a  tangent  and  a  chord  is 
measured  by  half  the  intercepted  arc. 

Let  the  tangent  B  E  form,  with 
the  chord  AC,  the  angle  BAC; 
then  BAC  is  measured  by  half 
the  arc  AMC. 

From  A,  the  point  of  contact, 
draw  the  diameter  AD.  The  an- 
gle BAD  is  a  right  angle  (Prop. 
X.),  and  is  measured  by  half  of 
the  semi-circumference  AMD 
(Prop.  XVIII.)  ;  and  the  angle  D  A  C  is  measured  by 
half  the  arc  D  C  ;  hence  the  sum  of  the  angles  BAD, 
D  A  C,  or  B  A  C,  is  measured  by  the  half  of  AMD,  plus 
the  half  of  D  C  ;  or  by  half  the  whole  arc  A  M  D  C. 

In  like  manner,  it  may  be  shown  that  the  angle  C  A  E 
is  measured  by  half  the  intercepted  arc  A  C. 


BOOK    III. 


75 


Proposition  XXI.  —  Theorem. 

207.    The  ang-le  formed  by  two  secants  is  measured  hy 
half  the  difference  of  the  tivo  intercepted  arcs. 

Let   A  B,  AC   be   two   secants  A 

forming  the  angle  BAG;  then 
will  that  angle  be  measured  by 
half  the  difference  of  the  two  arcs 
BEG  and  D  F. 

Draw  D  E  parallel  to  AG  ;  then 
will  the  arc  E  G  be  equal  to  the 
arc  D  F  (Prop.  XII.)  ;  and  the 
angle  B  D  E  be  equal  to  the  an- 
gle B  A  G  (Prop.  XXII.  Bk.  I.).  But  the  angle  B  D  E  is 
measured  by  half  the  arc  B  E  (Prop.  XVIII.)  ;  hence  the 
equal  angle  B  A  G  is  also  measured  by  half  the  arc  B  E  ; 
that  is,  by  half  the  difference  of  tlie  arcs  BEG  and  E  G, 
or,  since  E  G  is  equal  to  D  F,  by  half  the  differeucc  of  the 
intercepted  arcs  BEG  and  D  F. 


Proposition  XXII.  —  Theorem. 

208.  The  angle  formed  by  a  secant  and  a  tangent  is  meas- 
ured by  half  the  difference  of  the  tivo  intercepted  arcs. 

Let  the  secant  AB  form,  with  A 

the  tangent  A  G,  the  angle  BAG; 
then  BAG  is  measured  by  half 
the  difference  of  the  two  arcs 
BEF  and  FD. 

Draw  D  E  parallel  to  A  G  ;  then 
will  the  arc  E  F  be  equal  to  the  arc 
DF  (Prop.  XII.),  and  the  angle 
BDE  be  equal  to  the  angle  BAG. 
But  the  angle  B  D  E  is  measured  by  half  of  the  arc  B  E 
(Prop.  XVIII.)  ;  hence  the  equal  angle  BAG  is  also 
measured  ])y  lialf  the  arc  B  E ;  that  is,  by  lialf  the  difference 
of  the  arcs  P>EF  and  EF,  or,  since  EF  is  equal  to  DF,  by 
half  the  difference  of  tlie  intercepted  arcs  BEF  and  DF. 


BOOK    IV 


PROPORTIONS,   AREAS,   AND    SIMILARITY    OF 
FIGURES. 


DEFINITIONS. 

209.  The  area  of  a  figure  is  its  quantity  of  surface,  aud 
is  expressed  by  the  number  of  times  which  the  surface 
contains  some  other  area  assumed  as  a  unit  of  measure. 

Figures  have  equal  areas,  when  they  contain  the  same 
unit  of  measure  an  equal  number  of  times. 

210.  Similar  figures  are  such  as  have  the  angles  of 
the  one  eqvial  to  those  of  the  other,  each  to  each,  and  the 
sides  containing  the  equal  angles  proportional. 

211.  Equivalent  figures  are  such  as  have  equal  areas. 
Figures   may   be   equivalent   which    are    not    similar. 

Thus  a  circle  may  be  equivalent  to  a  square,  and  a  tri- 
angle to  a  rectangle. 

212.  Equal  figures  are  such  as,  when  applied  the  one 
to  the  other,  coincide  throughout  (Art.  84,  Ax.  14). 
Thus  circles  having  equal  radii  are  equal ;  and  triangles 
having  the  three  sides  of  the  one  equal  to  the  three  sides 
of  the  other,  each  to  each,  are  also  equal. 

Equal  figures  are  always  similar ;  but  similar  figures 
may  be  very  unequal. 

213.  In  diiferent  circles,  similar  arcs,  segments,  or 
sectors  are  such  as  correspond  to  equal  angles  at  the 
centres  of  the  circles. 


BOOK   IV. 


77 


Thus,  if  the  angles  A 
and  E  are  equal,  the  arc 
B  C  will  be  similar  to  the 
arc  FG;  the  segment  BDC 
to  the  segment  FHG,  and 
the  sector  ABC  to  the 
sector  EFG. 

214.  The   ALTITUDE  OF  A  TRIANGLE 

is  the  perpendicular,  which  measures 
the  distance  of  any  one  of  its  vertices 
from  the  opposite  side  taken  as  a 
base  ;  as  the  perpendicular  A  D  let 
fall  on  the  base  B  C  in  the  triangle 
ABC. 

215.  The  ALTITUDE  OF  A  PARALLEL- 
OGRAM is  the  perpendicular  which 
measures  the  distance  betw^een  its 
opposite  sides  taken  as  bases  ;  as  the 
perpendicular  E  F  measuring  the  dis- 
tance between  the  opposite  sides,  A  B 
allelogram  A  B  C  D. 

216.  The  ALTITUDE  OF  A  TRAPEZOID 

is  the  perpendicular  distance  between 
its  parallel-  sides  ;  as  the  distance 
measured  by  the  perpendicular  EF 
between  the  parallel  sides,  AB,  D  C, 
of  the  trapezoid  A  B  C  D. 

Proposition  I.  —  Theorem. 

217.  Parallelograms  which  have  equal  bases  and  equal 
altitudes  are  equivalent. 

Let  ABCD,  ABEF  be   two     D  C    F  E 

parallelograms  having  equal  bases 
and  equal  altitudes  ;  then  these 
paj-allelograms  are  equivalent. 

Let  the  base  of  the  one  paral- 

7* 


DC, 


F        B 

of  the  par- 


78  ELEMENTS   OF   GEOMETRY. 

lelogram  be  placed  on  that  of  the  other,  so  that  A  B  shall 
be  the  common  base.  Now,  since  the  two  parallelograms 
are  of  the  same  altitude,  their  upper  bases,  DC,  F E,  will 
be  in  the  same  straight  line,  D  C  E  F,  parallel  to  A  B. 
From  the  nature  of  parallelograms  D  C  is  equal  to  A  B, 
and  F  E  is  equal  to  A  B  (Prop.  XXXI.  Bk.  I.)  ;  therefore 
D  C  is  equal  to  F  E  (Art.  34,  Ax.  1)  ;  hence,  if  D  C  and 
F  E  be  taken  away  from  tiie  same  line,  D  E,  the  remain- 
ders C  E  and  D  F  will  be  equal  (Art.  34,  Ax.  3).  But 
A  D  is  equal  to  B  C  and  A  F  to  B  E  (Prop.  XXXI. 
Bk.  I.)  ;  therefore  the  triangles  D  A  F,  C  B  E,  are  mutu- 
ally equilateral,  and  consequently  equal  (Prop.  XVIII. 
Bk.  L). 

If  from  the  quadrilateral  ABED,  we  take  away  the  tri- 
angle A  D  F,  there  will  remain  the  parallelogram  ABEF; 
and  if  from  the  same  quadrilateral  A  B  E  D,  we  take  away 
the  triangle  C  B  E,  there  will  remain  the  parallelogram 
A  B  C  D.  Hence  the  parallelograms  A  BC  D,  ABE  F, 
which  have  equal  bases  and  equal  altitude,  are  equivalent. 

218.  Cor.  Any  parallelogram  is  equivalent  to  a  rec- 
tangle having  the  same  base  and  altitude. 

Proposition  II.  —  Theorem. 

219.  If  a  triang-le  and  a  parallelogram  have  the  same 
base  and  altitude^  the  triangle  is  equivalent  to  half  the 
parallelogram. 

Let  A  B  E  be  a  triangle,  and    D C    F  E 

A  B  C  D  a  parallelogram  having      V  V''       S^^^/ 

the   same   base,  A  B,  and   the        \       /0^^^\/ 
same   altitude ;    then   will   the  v>-^""^^    \x 

triangle  be   equivalent  to  half  A  B 

the  parallelogram. 

Draw  AF,  FE  so  as  to  form  the  parallelogram  ABEF. 
Then  the  parallelograms  A  B  C  D,  ABEF,  having  tlie 
same  base  and  altitude,  are  equivalent  (Prop.  I.).     But 


BOOK   IV.  •  79 

the  triangle  A  B  E  is  half  the  parallelogram  A  B  E  P 
(Prop.  XXXI.  Cor.  1,  Bk.  I.)  ;  hence  the  triangle  ABE 
is  equivalent  to  half  the  parallelogi*am  A  B  C  D  (Ai't.  34, 
Ax.  T). 

220.  Cor.  1.  Any  triangle  is  equivalent  to  half  a  rec- 
tangle having  the  same  base  and  altitude,  or  to  a  rectangle 
either  having  the  same  base  and  half  of  the  same  altitude, 
or  having  the  same  altitude  and  half  of  the  same  base. 

221.  Cor.  2.  All  triangles  which  have  equal  bases  and 
altitudes  are  equivalent. 

Proposition  III. — Theorejm. 

222.  Two  rectangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 

Let  ABCD,  AEFD  be     ^  F  C 

two  rectangles  having  tlie 
common  altitude  AD ;  they 
are  to  each  other  as  tlieir 
bases  A  B,  A  E.  A  E  B 

First.  Suppose  that  the  bases  AB,  AE  are  commensu- 
rable, and  are  to  each  other,  for  example,  as  the  numbers 
7  and  4.  If  A  B  is  divided  into  seven  equal  parts,  A  E 
will  contain  four  of  those  parts.  At  each  point  of  division 
draw  lines  perpendicular  to  the  base  ;  seven  rectangles 
will  thus  be  formed,  all  equal  to  each  other,  since  they 
have  equal  bases  and  the  same  altitude  (Prop.  I.). 
The  rectangle  A  B  C  D  will  contain  seven  partial  rectan- 
gles, while  A  E  F  D  will  contain  four ;  hence  the  rectangle 
A  B  C  D  is  to  A  E  F  D  as  7  is  to  4,  or  as  A  B  is  to  A  E. 
The  same  reasoning  may  be  applied,  whatever  be  the 
numbers  expressing  the  ratio  of  the  bases ;  hence,  what- 
ever be  that  ratio,  when  its  terms  are  commensurable,  we 
shall  have 

ABCD:AEFD::AB:AE. 


80  •      ELEMENTS  OF    GEOMETRY. 

Second.  Suppose  that  the  bases  A  B, 
A  E  are  incommensurable ;  we  shall 
still  have 

A  B  C  D  :  A  E  F  D  :  :  A  B  :  A  E. 
For,  if  this  proportion  be  not  true,  the     A  E  I  O  B 

first  three  terms  remaining  the  same,  the  fourth  term 
must  be  either  greater  or  less  than  A  E.  Suppose  it 
to  be  greater,  and  that  we  have 

ABCD:AEFD::AB:AO. 
Conceive  A  B  divided  hito  equal  parts,  each  of  which  is 
less  than  EG.  There  will  be  at  least  one  point  of  division, 
I,  between  E  and  0.  Through  this  point,  I,  draw  the 
perpendicular  I K  ;  then  the  bases  A  B,  A I  will  be  com- 
mensurable, and  we  shall  have 

ABCD  :  AIKD:  :  AB:  AI. 
But,  by  hypothesis,  we  have 

ABCD:AEFD::AB:AO. 

In  these  two  proportions  the  antecedents  are  equal ; 
hence  the  consequents  are  proportional  (Prop.  X.  Cor.  2, 
Bk.  11.),  and  we  have 

AIKD:AEFD::AI:AO. 

But  A  0  is  greater  than  A I ;  therefore,  if  this  proportion 
is  correct,  the  rectangle  A  E  F  D  must  be  greater  than  the 
rectangle  AIKD  (Art.  125)  ;  on  the  contrary,  however, 
it  is  less  (Art.  34,  Ax.  8)  ;  therefore  the  proportion  is 
impossible.  Hence,  ABCD  cannot  be  to  A  E  F  D  as 
A  B  is  to  a  line  greater  than  AE. 

In  the  same  manner,  it  may  be  shown  that  the  fourtli 
term  of  the  proportion  cannot  be  less  than  A  E  ;  therefore 
it  must  be  equal  to  AE.  Hoiice,  any  two  rectangles 
ABCD,  AEFD,  having  equal  altitudes,  are  to  each 
other  as  their  bases  A  B,  A  E. 


BOOK   IV.  81 


Proposition  IV.  —  Theorem. 

223.  Any  two  rectangles  are  to  eacn,  otJier  as  the  pro- 
ducts of  their  bases  multiplied  by  their  attitudes. 

Let  ABCD,  AEGF  be  two       n         d  C 

rectangles  ;  then  will  ABCD  be 
to  A  E  G  F  as  A  B  multiplied  by 
AD  is  to  A E  multiplied  by  A F.  E  - 
Having  placed  the  two  rectangles 
so  that  the  angles  at  A  arc  verti- 
cal, produce  the  sides  GE,  CD 


G 


till  they  meet  iji  H.  The  two  rectangles  ABCD, 
AEHD,  having  the  same  altitude,  AD,  are  to  each  other 
as  their  bases,  A  B,  AE.  In  like  manner  the  two  rectan- 
gles AEHD,  AEGF,  having  the  same  altitude,  A  E,  are 
to  each  other  as  their  bases,  AD,  AF.  Heuce  wc  have 
the  two  proportions, 

A  B  C  D  :  A  E  H  D  :  :  A  B  :  A  E, 
AEHD  :  AEGF:  :  AD  :  AF. 

Multiplying  the  corresponding  terms  of  these  propor- 
tions together  (Prop.  XIH.  Bk.  H.),  and  omitting  the 
factor  AEHD,  which  is  common  to  both  the  antecedent 
and  the  consequent  (Prop.  IX.  Bk.  II.),  we  shall  have 

ABCD:AEGF::ABXAD:AEXAF. 

224.  Scholium.  Hence,  we  may  assume  as  tlie  measure 
of  a  rectangle,  the  product  of  its  base  by  its  altitude,  pro- 
vided we  understand  by  this  product  the  product  of  two 
numbers,  one  of  which  represents  the  number  of  linear 
units  contained  in  the  base,  the  other  the  number  of  linear 
units  contained  in  the  altitude. 

The  product  of  two  lines  is  often  used  to  designate 
their  rectangle ;  but  the  term  square  is  used  to  designate 
the  product  of  a  number  multiplied  by  itself. 


82 


ELEMENTS    OF   GEOMETRY. 


Proposition  V.  —  Theorem. 

225.  The  area  of  any  parallelogram  is  equal  to  the  pro- 
duct of  its  base  by  its  altitude. 

Let  A  B  C  D  be  any  parallelogram, 
A  B  its  base,  and  B  E  its  altitude  ; 
then  will  its  area  be  equal  to  the  pro- 
duct of  A  B  by  B  E. 

Draw  BE  and  AF  perpendicular     A  B 

to  A B,  and  produce  CD  to  F.  Then  the  parallelogram 
A  B  C  D  is  equivalent  to  the  rectangle  A  B  E  F,  which  has 
the  same  base,  AB,  and  the  same  altitude,  BE  (Prop.  I. 
Cor.).  But  the  rectangle  A B E F  is  measured  by  A  B  X 
BE  (Prop.  lY.  Sch.)  ;  therefore  AB  X  BE  is  equal  to 
the  area  of  the  parallelogram  A  B  C  D. 

226.  Cor.  Parallelograms  having  equal  bases  are  to 
each  other  as  their  altitudes,  and  parallelograms  having 
equal  altitudes  are  to  each  other  as  their  bases ;  and,  in 
general,  parallelograms  are  to  each  other  as  the  products 
of  their  bases  by  their  altitudes. 


E 


Proposition  YI.  —  Theorem. 

227.    The  area  of  any  triangle  is  equal  to  the  product 
of  its  base  by  half  its  altitude 

Let  A  B  C  be  any  triangle,  B  C  its  base, 
and  A  D  its  altitude  ;  then  its  area  will  be 
equal  to  the  product  of  B  C  by  half  of  A  D. 

Draw  A  E  and  C  E  so  as  to  form  the 
parallelogram  A  B  C  E  ;  then  the  triangle 
A  B  C  is  half  the  parallelogram  A  B  C  E, 
wliich  has  the  same  base  B  C,  and  the  same  altitude  A  D 
(Prop.  II.)  ;  but  the  area  of  the  parallelogram  is  equal  to 
B  C  X  A  D  (Prop.  Y.)  ;  hence  the  area  of  the  triangle 
must  be  ^  BC  X  AD,  or  B  C  X  |  AD. 


BOOK   IV.  83 

228.  Cor,  Triangles  of  equal  altitudes  are  to  each 
other  as  their  bases,  and  triangles  of  equal  bases  are  to 
each  other  as  their  altitudes  ;  and,  in  general,  triangles 
are  to  each  other  as  the  products  of  their  bases  and  alti- 
tudes. 

Proposition  YII.  —  Theorem. 

229.  The  area  of  any  trapezoid  is  equal  to  the  product 
of  its  altitude  by  half  the  sum  of  its  parallel  sides. 

Let  A  B  C  D  be  a  trapezoid,  E  F 
its  altitude,  and  A B,  CD  its  par- 
allel sides  ;  then  its  area  will  be 
equal  to  the  product  of  E  F  by  half 
the  sum  of  A  B  and  C  D. 


Through  I,  the  middle  point  of    A  F         LB 

the  side  B  C,  draw  K  L  parallel  to  AD  ;  and  produce  DO 
till  it  meet  K  L.  In  the  triangles  I  B  L,  I  C  K,  we  have 
the  sides  I B,  I  C  equal,  by  construction  ;  the  vertical  an- 
gles LIB,  C I K  are  equal  (Prop.  IV.  Bk.  I.)  ;  and,  since 
CK  and  BL  are  parallel,  the  alternate  angles  IBL,  ICK 
are  also  equal  (Prop.  XXII.  Bk.  I.)  ;  therefore  tlie  trian- 
gles I B  L,  I  C  K  are  equal  (Prop.  YI.  Bk.  I.)  ;  hence  the 
trapezoid  ABCD  is  equivalent  to  the  parallelogram 
A  D  K  L,  and  is  measured  by  the  product  of  E  F  by  A  L 
(Prop.  v.). 

But  we  have  AL  equal  DK;  and  since  the  triangles 
IBL  and  K  C  I  are  equal,  the  sides  B  L  and  C  K  are 
equal ;  therefore  the  sum  of  A  B  and  C  D  is  equal  to  the 
sum  of  A  L  and  D  K,  or  twice  A  L.  Hence  A  L  is  lialf 
the  sum  of  the  bases  A  B,  CD;  hence  the  area  of  the 
trapezoid  A  B,  C  D  is  equal  to  the  product  of  the  altitude 
E  F  by  half  the  sum  of  the  parallel  sides  AB,  CD. 

Cor.  If  through  I,  the  middle  point  of  B  C,  the  line  IH 
be  drawn  parallel  to  the  base  AB,  the  point  H  will  also 
be  the  middle  point  of  AD.     For,  since  the  figure  AH  IL 


84  ELEMENTS   OF   GEOMETRY. 

is  a  parallelogram,  as  is  likewise  D  H I K,  their  opposite  sides 
being  parallel,  we  have  A  H  equal  to  I L,  and  D  H  equal 
to  IK.  But  since  the  triangles  BIL,  CIK  are  equal, 
we  have  I  L  equal  to  I  K  ;  hence  A  H  is  equal  to  D  H. 

Now,  the  line  H  I  is  equal  to  A  L,  which  has  been 
shown  to  be  equal  to  half  tlie  sum  of  A  B  and  C  D  ;  there- 
fore the  area  of  the  trapezoid  is  equal  to  the  product  of 
E F  by  HI.  Hence,  the  area  of  a  trapezoid  is  equal  to 
the  product  of  its  altitude  by  the  line  connecting  tlie  mid- 
dle points  of  tlie  sides  wliich  are  not  parallel. 

Proposition  VIII.  —  Theorem. 

230.  If  a  straight  line  he  divided  into  tivo  parts,  the 
sqvare  described  on  the  v^hole  line  is  equivalent  to  the 
sum  of  the  squares  described  on  the  parts,  together  with 
twice  the  rectangle  contained,  by  the  parts. 

Let  AC  be  a  straight  line,  divided       e  H     D 

into  two  parts,  AB,  BC,  at  the  point  B; 

then   the   square    described   on   AC  is      F _j G 

equivalent  to  the  sum  of  the  squares 
described  on  the  parts  AB,  B  C,  togetli- 
er  with  twice  the  rectangle  contained 
by  A  B,  B  C  ;  that  is, 

AC^  =  AB^  4-  B~C^  +  2  AB  X  B  C. 

On  A  C  describe  the  square  A  C  D  E  ;  take  AF  equal  to 
A  B  ;  draw  F  G  parallel  to  A  C,  and  B  H  parallel  to  A  E. 

The  square  A  C  D  E  is  divided  into  four  parts;  the  first, 
A  B I F,  is  the  square  described  on  A  B,  since  A  F  was 
taken  equal  to  A  B.  The  second,  I  G  D  H,  is  the  square 
described  upon  B  C  ;  for,  since  A  C  is  equal  to  A  E,  and 
A  B  is  equal  to  AF,  AC  minus  AB  is  equal  to  AE  minus 
A  F,  which  gives  B  C  equal  to  E  F.  But  I  G  is  equal  to 
B  C,  and  D  G  to  EF,  since  tlie  lines  are  parallels  ;  there- 
fore I  G  D  H  is  equal  to  the  square  described  on  B  C. 


I 

B     C 


BOOK   IV. 


85 


These  two  parts  being  taken  from  the  whole  square,  there 
remain  two  rectangles  B  C  G  I,  E  F  I  H,  eacli  of  which  is 
measured  by  A  B  X  B  C  ;  hence  the  square  on  the  whole 
line  AC  is  equivalent  to  the  squares  on  the  parts  AB,  BC, 
together  with  twice  the  rectangle  of  the  parts. 

E        H         D 


281.  Cor.  The  square  described  on 
the  whole  line  A  C  is  equivalent  to  four 
times  the  square  described  on  the  half 
AB. 


1 

G 


B 


232.   Scholium.     This  proposition  is  equivalent  to  the 
algebraical  formula, 


L     F 


K 


H 


Proposition  IX.  —  Theorem. 

233.  The  square  described  on  the  difference  of  tivo 
straight  lines  is  equivalent  to  the  sum  of  the  squares  de- 
scribed on  the  tivo  lines,  diminished  by  twice  the  rectangle 
contained  by  the  lines. 

Let  AB  and  BC  be  two  lines,  and 
A  C  their  difference  ;  then  will  the 
square  described  on  A  C  be  equiva- 
lent to  the  sum  of  the  squares  de- 
scribed on  AB,  B  C,  diminished  by 
twice  the  rectangle  AB,  B  C ;  that  is,  a  C     B 

(AB  —  B  Cy  or  AC'  =  AB'  +  FC^  —  2  A  B  X  BC. 

On  A  B  describe  the  square  A  B I F  ;  take  A  E  equal  to 
A  C  ;  draw  C  Gr  parallel  to  B  I,  H  K  parallel  to  AB,  and 
complete  the  square  E  F  L  K. 

Since  AF  is  equal  to  AB,  and  AE  to  AC,  EF  is  equal 
to  B  C,  and  L  F  to  G I ;  therefore  L  G  is  equal  to  F I ; 
hence   the   two   rectangles    CBIG,    GLKD    are    each 


E 

D 

86  ELEMENTS   OP   GEOMETRY. 

measured  by  A  B  X  B  C.  Take  these  rectangles  from 
the  whole  figure  ABILKE,  which  is  equivalent  to 
A  B2  -f-  B  C\  and  there  will  evidently  remain  the  square 
A  C  D  E ;  hence  the  square  on  A  C  is  equivalent  to  the 
sum  of  the  squares  on  A  B,  B  C,  diminished  by  twice  the 
rectangle  contained  by  AB,  B  C. 

234.  Scholium.  This  proposition  is  equivalent  to  the 
algebraical  formula, 

Proposition  X.  —  Theorem. 

235.  The  rectangle  contained  by  the  sum  and  difference 
of  two  straight  lines  is  equivalent  to  the  difference  of  the 
squares  of  these  lines. 

F  G     I 

Let  A  B,  B  C  be  two  lines ;  then 
will  the  rectangle  contained  by  the 
sum  and  difference  of  A  B,  B  C,  be 
equivalent  to  the  difference  of  the 
squares  of  A  B,  B  C  ;  that  is,  A  C     B     K 

(AB  +  BC)  X  (AB  — BC)  =  AB'— BCl 

On  AB  describe  the  square  ABIF,  and  on  AC  the 
square  A  C  D  E  ;  produce  CD  to  G  ;  and  produce  A  B 
until  B  K  is  equal  to  B  C,  and  complete  the  rectangle 
AKLE. 

The  base  A  K  of  the  rectangle  is  the  sum  of  the  two 
lines  A  B,  B  C  ;  and  its  altitude  A  E  is  tlie  difference  of 
the  same  lines  ;  therefore  the  rectangle  AKLE  is  that 
contained  by  the  sum  and  the  dififercnco  of  the  lines  A  B, 
B  C.  But  this  rectangle  is  composed  of  the  two  parts 
A  B  H  E  and  B  H  L  K  ;  and  the  part  B  H  L  K  is  equal 
to  the  rectangle  E  D  GrF,  since  B  H  is  equal  to  D  E,  and 
BK  to  EF.  Hence  the  rectangle  AKLE  is  equivalent 
to  A  B  H  E  plus  EDGE,  which  is  equivalent  to  the  dif- 


H 

D 

BOOK   IV. 


87 


fereiice  between  the  square  A  B I F  described  on  A  B,  and 
D  H I  Gr  described  on  B  C  ;  hence 


AB^ 


BC\ 


(AB  +  BC)  X  (AB  — BC) 

236.   Scholium.   This  proposition  is  equivalent  to  the 
algebraical  formula, 

(a  +  b)  X  Ca  —  b)  =  a''  —  b\ 


Proposition  XI.  —  Theorem. 

237.  The  square  described  on  the  hypothenuse  of  a 
right-angled  triangle  is  equivalent  to  the  sum  of  the 
squares  described  on  the  other  tivo  sides. 

Let  ABC  be  a  right-angled  L 

triangle,  having  the  right  angle 
at  A;  then  the  square  described  H, 
on  the  hypothenuse  B  C  will  be 
equivalent  to  the  sum  of  the 
squares  on  the  sides  BA,  AC. 

On  B  C  describe  the  square 
BCGF,  andon  AB,  AC  the 
squares  ABHL,  ACIK;  and 
through  A  draw  AE  parallel  to 
BFor  CG,  and  join  AF,HC. 

The  angle  A  B  F  is  composed  of  the  angle  A  B  C,  to- 
gether with  the  right  angle  C  B  F  ;  the  angle  C  B  H  is 
composed  of  the  same  angle  ABC  together  with  the  right 
angle  A  B  H ;  therefore  the  angle  A  B  F  is  equal  to  the 
angle  H  B  C.  But  we  have  A  B  equal  to  B  H,  being  sides 
of  the  same  square  ;  and  B  F  equal  to  B  C,  for  the  same 
reason  ;  therefore  the  triangles  A  B  F,  H  B  C  liave  two 
sides  and  the  included  angle  of  the  one  equal  to  two 
sides  and  the  included  angle  of  the  other ;  hence  they  are 
themselves  equal  (Prop.  Y.  Bk.  I.). 

But  the  triangle  A  B  F  is  equivalent  to  half  the  rectan- 
gle B  D  E  F,  since  they  have  the  same  base  B  F,  and  the 


E    G 


88 


ELEMENTS   OP   GEOMETRY. 


same  altitude  BD  (Prop.  II. 
Cor.  1).  The  triangle  HBC 
is,  in  like  manner,  equivalent  H 
to  half  the  square  A  B  H  L ;  for 
the  angles  BAG,  BAL  being 
both  right,  AC  and  AL  form 
one  and  the  same  straiglit  line 
parallel  to  HB  (Prop.  II.  Bk. 
I.)  ;  and  consequently  the  tri- 
angle and  the  square  have  the 
same     altitude     A  B      (Prop.  ^  ^    ^ 

XXy.  Bk.  I.)  ;  and  they  also  have  the  same  base  B  H ; 
hence  the  triangle  is  equivalent  to  half  the  square  (Prop. 

II.). 

The  triangle  A  B  F  has  already  been  proved  equal  to  the 
triangle  HBC;  hence  the  rectangle  BDEF,  which  is 
double  the  triangle  A  B  F,  must  be  equivalent  to  the 
square  ABHL,  which  is  double  the  triangle  HBC.  In 
the  same  manner  it  may  be  proved  that  the  rectangle 
C  D  E  G  is  equivalent  to  the  square  A  C  I  K.  But  the 
two  rectangles  BDEF,  CDEG,  taken  together,  compose 
the  square  BCGF;  therefore  the  square  B  C  G  F,  de- 
scribed on  the  hypotlienuse,  is  equivalent  to  the  sum  of 
the  squares  ABHL,  ACIK,  described  on  the  two  other 
sides ;  that  is, 

FC^  is  equivalent  to  AB^  +  AC^- 

238.  Cor.  1.  The  square  of  either  of  the  sides  which 
form  the  right  angle  of  a  right-angled  triangle  is  equiva- 
lent to  the  square  of  the  hypothenuse  diminished  by  the 
square  of  the  other  side ;  thus, 

A  B^  is  equivalent  to  B  C  —  A  C^. 

239.  Cor.  2.  The  square  of  the  hypothenuse  is  to  the 
square  of  either  of  the  other  sides,  as  the  hypothenuse  is 
to  the  part  of  the  hypothenuse  cut  off,  adjacent  to  that  side, 


BOOK   IV.  89 

hy  the  perpendicular  let  fall  from  the  vertex  of  the  rig'ht 
angle.  For,  on  account  of  the  common  altitude  B  F,  the 
square  BCGF  is  to  the  rectangle  BDEF  as  the  base  BC 
is  to  the  base  B  D  (Prop.  III.)  ;  now,  the  square  A  B  H  L 
has  been  proved  to  be  equivalent  to  the  rectangle  BDEF; 
therefore  we  have, 

BC^  AB^  :BC:BD. 

In  like  manner,  we  have, 

W&  :  AC^  :  :  B C  :  C D. 

240.  Cor,  ^.  If  a  perpendicular  he  drawn  from  the 
vertex  of  the  right  angle  to  the  hf/pothenuse,  the  squares 
of  the  sides  about  the  right  angle  will  be  to  each  other  as 
the  adjacent  segments  of  the  hypothenuse.  For  the  rec- 
tangles BDEF,  DC GE,  having  the  same  altitude,  are  to 
each  other  as  their  bases,  B  D,  C  D  (Prop.  III.).  But 
these  rectangles  are  equivalent  to  the  squares  A  B  H  L, 
A  C  I K ;  therefore  we  have, 

AB"  :  AC^  :  :  B  D  :  D  C. 

241.  Cor,  4.  The  square  described  on 
the  diagonal  of  a  square  is  equivalent  to 
double  the  square  described  on  a  side. 
For  let  A  B  C  D  be  a  square,  and  A  C 
its  diagonal ;  the  triangle  ABC  being 
right-angled  and  isosceles,  we  have. 


B 


AC"=  AB'-f  BC"=2AB^=2  X  ABCD. 

242.   Cor.  5.    Since  A  C^  is  equal  to  2  A  B^,  we  have 
AC^  :  AB'  :  :  2  :  1 ; 
and,  extracting  the  square  root,  we  have 
AC  :  AB  :  :  V2:  1; 

hence,  the  diagonal  of  a  square  is  incommensurable  with 

a  side. 

8* 


90 


ELEMENTS   OP   GEOMETRY. 


M 


243.  Note. — The  proposition  may  also  be  demonstrated 
as  follows  :  — 

Let  ABC  be  a  right-angled 
triangle,  having  the  right  angle 
at  A ;  then  the  square  described 
on  the  hypothenuse  B  C  will 
be  equivalent  to  the  sum  of 
tlie  squares  on  the  sides  BA, 
AC. 

On  B  C  describe  the  square 
BCGF,  and  on  AB,  ACthe 
squares  ABHL,  AC  IK;  pro- 
duce F  B  to  N,  HL  £nd  I K  to 
M ;  and  through  A  draw  EDA 
parallel  to  FBN,  and  meeting 
the  prolongation  of  H  L  in  M. 

Then,  since  the  angles  H  B  A,  N  B  C  are  both  right  an- 
gles, if  the  common  angle  N  B  A  be  taken  from  each  of 
these  cquxils,  there  will  remain  the  equal  angles  H  B  N, 
ABC;  and,  consequently,  since  the  triangles  H  B  N, 
ABC  are  both  right-angled,  and  have  also  the  sides  B  H, 
B  A  equal,  their  hypothenuses  B  N,  B  C  are  equal  (Prop. 
yi.  Cor.,  Bk.  I.).  But  B  C  is  equal  to  B  F  ;  therefore 
B  N  is  equal  to  B  F  ;  hence  the  parallelograms  B  A  M  N, 
B  D  E  F,  of  which  the  common  altitude  is  B  D,  have  equal 
bases ;  therefore  the  two  parallelograms  are  equivalent 
(Prop.  I.).  But  the  parallelogram  B  AM  N  is  equivalent 
to  the  square  ABHL,  since  they  have  the  same  base 
B  A,  and  the  same  altitude  A  L ;  hence  the  parallelogram 
B D  E F  is  also  equivalent  to  the  square  ABHL.  In  like 
manner  it  may  be  shown  that  the  rectangle  D  C  G  E  is 
equivalent  to  the  square  A  C  I K  ;  hence  the  two  rectan- 
gles together,  that  is,  the  square  BCGF,  are  equivalent 
to  the  sum  of  the  squares  ABHL,  A  C I K. 


BOOK   IV.  91 


Proposition  XII.  —  Theorem. 

244.  In  any  triangle,  the  square  of  the  side  opposite  an 
acute  angle  is  less  than  the  sum  of  the  squares  of  the  base 
and  the  other  side,  by  tiuice  the  rectangle  contained  by  the 
base  and  the  distance  from  the  vertex  of  the  acute  angle 
to  the  perpendicular  let  fall  from  the  vertex  of  the  opposite 
angle  on  the  base,  or  on  the  base  produced. 

Let  ABC  be  any  triangle,  A  A 

C  one  of  its  acute  angles,  /K  jK 

and  AD  the  perpendicular  /    I  \  l\\ 

let  fall  on  the  base  B  C,  or         /     i     \  \    \ 

on  B  C  produced ;  then,  in       /       j       \  \      \ 

cither  case,  will  the  square      ^ ^ -^       il"  "p p 

of  A  B  be  less  than  the  sum 

of  the  squares  of  AC,  B  C,  by  twice  the  rectangle  B  C  X 

CD. 

First.  Wlien  the  perpendicular  falls  within  the  triangle 
ABC,  we  have   BD  =  BC  —  CD;  and  consequently, 
BD'  =  BC^  +  CD^  —  2  B  C  X  C  D  (Prop.  IX.).     By 
adding  A  D   to  each  of  tliese  equals,  we  have 
BD'  +  ad'  =  BC'  +  CD'  +  AD'  _  2  BC  X  C D. 
But  the  two  right-angled  triangles  A  D  B,  ADC  give 

AB'  =  BD'  +  AD',    and    AC'  =  CD'  +  AD' 
(Prop.  XI.)  ;  therefore, 

AB'  =  BC'  +  AC'  —  2  B C  X  C D. 

Secondly.  When  the  perpendicular  A  D  falls  without  the 
triangle  ABC,  we  have  BD  =  CD  —  BC;  and  conse- 
quently, B  D'  ==  CD'  +  BC'  _  2  C  D  X  B  C.  By  add- 
ing A  D   to  each  of  these  equals,  we  find,  as  before, 

AB'  =  B  C'  +  AC'  —  2  B C  X  C D. 


92  ELEMENTS    OF   GEOMETRY. 


Proposition  XIII.  —  Theorem. 

245.  In  any  ohtuse-angled  triangle^  the  square  of  the 
side  opposite  the  obtuse  angle  is  equivalent  to  the  sum  of 
the  squares  of  the  two  other  sides  plus  twice  the  rectangle 
contained  by  the  one  of  those  sides  into  the  distance  from 
the  vertex  of  the  obtuse  angle  to  the  perpendicular  let  fall 
from  the  vertex  of  the  opposite  angle  to  that  side  produced. 

Let  A  C  B  be  an  obtuse-angled  triangle,  A 
having  the  obtuse  angle  at  C,  and  let  A  D 
be  perpendicular  to  the  base  B  C  produced  ; 
then  the  square  of  A  B  is  greater  than  the 
sum  of  the  squares  of  B  C,  AC,  by  twice 
the  rectangle  B  C  X  C  D.  Since  B  D  is  the 
suna  of  the  lines  B  C  +  C  D,  we  have 

B"D^  =  Blf  +  CD'  +  2  BC  X  CD 

(Prop.  VIII.).     By  adding  AD    to  each  of  these  equals, 
we  have 

BD'  +  AD'=B"C'+CD'  + AD'  +  2BC  X  CD. 
But  the  two  right-angled  triangles  A  D  B,  A  D  C  give 

AB'  =  BD'  +  AD^  .   and     AC'  =  CD'  +  AD^ 
(Prop.  XI.)  ;  therefore, 

AB'  =  BC'  +  AC'-|-2BC  X  CD. 

246.  Scholium.  The  right-angled  triangle  is  the  only 
one  in  which  the  sum  of  the  squares  of  two  sides  is  equiv- 
alent to  the  square  of  the  third ;  for  if  the  angle  contained 
by  the  two  sides  is  acute,  the  sum  of  their  squares  will  be 
greater  than  the  square  of  the  opposite  side  ;  if  obtuse,  it 
will  be  less. 

Proposition  XIY.  —  Theorem. 

247.  In  any  triangle,  if  a  straight  line  be  drawn  from 
the  vertex  to  the  middle  point  of  the  base,  the  sum  of  the 


BOOK   IV. 


93 


squares  of  the  other  two  sides  is  equivalent  to  twice  the 
square  of  the  bisecting  line,  together  with  twice  the  square 
of  half  the  base. 

In  any  triangle  ABC,  draw  the  line  A 

A  E  from  the  vertex  A  to  the  middle  of 
the  base  B  C ;  then  the  sum  of  the  squares 
of  the  two  sides,  A  B,  AC,  is  equivalent 
to  twice  the  square  of  A  E  together  with 
twice  the  square  of  B  E. 

On  B  C  let  fall  the  perpendicular  AD  ;  ^ 
then,  in  the  triangle  ABE, 

AB^  =  AE^  +  EB^  +  2  E  B  X  E  D 
(Prop.  XIII.),  and,  in  triangle  AE  C, 

AC^  =  AE^  +  W(f  —  2  E  C  X  E  D 
(Prop.  XII.).     Hence,  by  adding  the  corresponding  sides 
together,  observing  that  since  E  B  and   E  C  are  equal, 

EB^  is  equal  to  E~C^  and  EBxEDtoECxED,  we 
have 

AB^  +  A C^  =  2  AE^  +  2  EB^*- 


Proposition  XV.  —  Theorem. 

248.  In  any  parallelogram  the  sum  of  the  squares  of 
the  four  sides  is  equivalent  to  the  sum  of  the  squares  of 
the  tivo  diagonals. 

Let  A  B  C  D  be  any  parallelogram,  d  q 

the  diagonals  of  which  are  AC,  BD ; 
then  the  sum  of  the  squares  of  A  B, 
B C,  CD,  DA  is  equivalent  to  the 
sum  of  the  squares  of  A  C,  B  D. 

For  the  diagonals  AC,  B D  bisect 
each  other  (Prop.  XXXIV.  Bk.  I.)  ; 
gle  ABC,  Xb'  +  B^'  ==   2  AE' 
XIV.)  ;  also,  in  the  triangle  A  D  C, 


hence,  in  the  trian- 
+  2  be'    (Prop. 


AD'  +  D  C'  =  2  A  E'  +  2  D  E^ 


94  ELEMENTS    OF   GEOMETRY. 

Hence,  by  adding  the  corresponding  sides  together,  and 
observing  that,  since  B  E  and  D  E  are  equal,  B  E^  and  D  E^ 
must  also  be  equal,  we  shall  have, 

AB'  +  bo'  +  ad'  +  DC'  =  4  AE'  +  4  DE'. 


But  4  A  E  is  the  square  of  2  A  E,  or  of  A  C,  and  4  D  E' 
is  the  square  of  2  DE,  or  of  BD  (Prop.  YIII.  Cor.); 
hence, 

BA'  +  BC'  +  C'D'  +  ad'  =  AC'  +  BD'. 

Proposition  XVI.  —  Theorem. 

249.  In  any  quadrilateral  the  sum  of  the  squares  of  the 
sides  is  equivalent  to  the  sum  of  the  squares  of  the  diag- 
onals^ plus  four  times  the  square  of  the  straight  line  that 
joins  the  middle  points  of  the  diagonals. 

Let  A  B  C  D  be  any  quadrilateral,  the 
diagonals  of  which  are  AC,  D  B,  and 
E  P  a  straight  line  joining  their  mid- 
dle points,  E,  F;  then  the  sum  of  the 
squares  of  AB,  B  C,  C  D,  AD  is  equiv- 
alent to  AC'  +  BD'  +  4  EP'.  A 

Join  E  B  and  E  D  ;  then  in  the  triangle  ABC, 

AB'  +  BC'  =  2  AE'  +  2  BE^ 
(Prop.  XIY.),  and  in  the  triangle  ADC, 

AD'  +  CD'  =  2  AE'  -f  2  DE'. 
Hence,  by  adding  the  corresponding  sides,  we  have 

AB' +  BC' +  AD'+ CD'=:  4  AE'  + 2  5^+ 2  DE'. 
But  4  AE'  is  equivalent  to  AC'  (Prop.  VHI.  Cor.),  and 
2  BT]'  +  2  D"e'  is  equivalent  to  4  B~p'  +  4  EP'  (Prop. 
XIV.);   hence, 

Al3'  +  BC'  -f  ad'  +  C^'  =  AC'  +  BD^  +  4  EP'. 


BOOK  IV.  95 

250.  Cor.  If  the  quadrilateral  is  a  parallelogram,  the 
points  E  and  F  will  coincide  ;  then  the  proposition  will  be 
the  same  as  Prop.  XY. 

251.  Scholium.  Proposition  XY.  is  only  a  particular 
case  of  this  proposition. 

Proposition  XYII.  —  Theorem. 

252.  If  a  straight  line  be  draivn  in  a  triangle  parallel 
to  one  of  the  sides,  it  vnll  divide  the  other  tivo  sides  pro- 
portionally. 

Let  ABC  be  a  triangle,  and  D  E  a 
straiglit  line  drawn  within  it  parallel  to 
the  side  B  C  ;  then  will 

AD  :  DB::  AE:EC. 

Join  B  E  and  D  C  ;  then  the  two  trian- 
gles B  D  E,  DEC  have  the  same  base, 
D  E ;  they  have  also  the  same  altitude, 
since  the  vertices  B  and  C  lie  in  a  line  parallel  to  the  base ; 
therefore  the  triangles  are  equivalent  (Prop.  11.  Cor.  2). 

The  triangles  A  D  E,  B  D  E,  having  their  bases  in  the 
same  line  A  B,  and  having  the  common  vertex  E,  have  the 
same  altitude,  and  therefore  are  to  each  other  as  their 
bases  (Prop.  YI.  Cor.)  ;  hence 

ADE:BDE::AD:DB. 

The  triangles  A  D  E,  D  E  C,  whose  common  vertex  is  D, 
liave  also  tlie  same  altitude,  and  therefore  are  to  each 
other  as  their  bases  ;  hence 

ADE:DEC::AE:EC. 

But  the  triangles  B  D  E,  DEC  have  been  sliown  to  be 
equivalent ;  therefore,  on  account  of  the  common  ratio  in 
the  two  proportions  (Prop.  X.  Bk.  II.), 

A  D  :  D  B  :  :  A  E  :  E  C. 

253.  Cor.  1.   Hence,  by  composition  (Prop.  YII.  Bk. 


96 


ELEMENTS   OF   GEOMETRY. 


IL), 
AB 


we  have  AD  +  D  B  :  AD  :  :  AE  +  E  C  :  AE,  or 
AD  :  :  AC  :  AE;  also,  AB:  BD  :  :  AC  :EC. 


254.  Co7\  2.  If  two  or  more  straight  lines  be  drawn  in 
a  triangle  parallel  to  one  of  the  sides,  they  will  divide  the 
other  two  sides  proportionally. 

For,  in  the  triangle  ABC,  since  D  E  ♦  A 

is  parallel  to  B  C,  by  the  theorem,  A  D  : 
D  B  :  :  A  E  :  E  C  ;  and,  in  the  triangle 
A  D  E,  since  F  G  is  parallel  to  D  E,  by 
the  preceding  corollary,  AD  :  FD  :  : 
A  E  :  G  E.  Hence,  since  the  antece- 
dents are  the  same  in  the  two  propor- 
tions (Prop.  X.  Cor.  2,  Bk.  II.),  F  D  :  D  B  :  :  G  E  :  E  C. 


Proposition   XVIII.  —  Theorem. 

255.  If  a  straight  line  divides  tivo  sides  of  a  triangle 
proportionally^  the  line  is  parallel  to  the  other  side  of  the 
triangle. 

Let  ABC  be  a  triangle,  and  D  E  a  ^ 

straight  line  drawn  in  it  dividing  the 
sides  AB,  A  C,  so  that  AD  :  D  B  :  :  AE  : 
E  C ;  then  will  the  line  D  E  be  parallel 
to  the  side  B  C. 

Join  B  E  and  D  C  ;  then  the  triangles 
A  D  E,  B  D  E,  having  their  bases  in  the 
same  straight  line  AB,  and  having  a  common  vertex,  E, 


are  to  each  other  as 
Cor.)  ;  that  is, 

ADE 


their  bases  AD,  DB    (Prop.  VI. 


BDE:  :  AD  :  DB. 


Also,  the  triangles  ADE,  DEC,  having  the  common 
vertex  D,  and  their  bases  in  the  same  line,  are  to  each 
other  as  these  bases,  A  E,  EC;  that  is, 

A  D  E  :  D  E  C  :  :  A  E  :  E  C. 


BOOK  IV.  97 

But,  by  hypothesis,  A  D  :  D  B  :  :  A  E  :  E  C ;  hence  (Prop. 
X.  Bk.  II.), 

ADE:BDE::ADE:DEC; 

that  is,  BDE,  DE  C  have  the  same  ratio  to  ADE  ;  there- 
fore the  triangles  BDE,  DEC  have  the  same  area,  and. 
consequently  are  equivalent  (Art.  211).  Since  these  tri- 
angles have  the  same  base,  D  E,  their  altitudes  are  equal 
(Prop.  VI.  Cor.)  ;  hence  the  line  B  C,  in  which  their  ver- 
tices are,  must  be  parallel  to  D  E. 

Proposition  XIX.  —  Theorem. 

256.  The  straight  line  bisecting'  any  angle  of  a  triangle 
divides  the  opposite  side  into  parts,  which  are  proportional 
to  the  adjacent  sides. 

In  any  triangle,  ABC,  let  the  an-     E 
gle  B  A  C  be  bisected  by  the  straight       (■■••..... 
line  A  D  ;  then  will 

BD  :  DC::AB:AC. 

Through   the   point   C   draw  C  E 
parallel   to   AD,  meeting  BA   pro-        ^        ^  ■" 

duced  in  E.  Then,  since  the  two  parallels  AD,  EC 
are  met  by  the  straight  line  A  C,  the  alternate  angles 
D  A  C,  ACE  are  equal  (Prop.  XXII.  Bk.  I.)  ;  and  the 
same  parallels  being  met  by  the  straight  line  B  E,  the  op- 
posite exterior  and  interior  angles  BAD,  A  E  C  are  also 
equal  (Prop.  XXII.  Bk.  I.).  But,  by  hypothesis,  tlie  an- 
gles DAC,  BAD  are  equal;  consequently  the  angle  ACE 
is  equal  to  the  angle  A  E  C  ;  hence  the  triangle  A  C  E  is 
isosceles,  and  the  side  AE  is  equal  to  the  side  A  C  (Prop. 
VIII.  Bk.  I.).  Again,  since  A  D,  in  the  triangle  E  B  C, 
is  parallel  to  E  C,  we  have  BD:DC::AB:AE  (Prop. 
XVII.),  and,  substituting  A  C  in  place  of  its  equal  A  E, 

BD:DC::AB:AC. 


98  ELEMENTS    OF   GEOMETRY. 


Proposition  XX.  —  Theorem. 

257.  If  a  straight  line  draiim  from  the  vertex  of  any 
angle  of  a  triangle  divides  the  opposite  side  into  parts 
tvhich  are  proportional  to  the  adjacent  sides,  the  line  bi- 
sects the  angle. 

Let  the  straight  line  AD,  drawn      E 
from  the  vertex  of  the  angle  BAG,       f  •.. 
in  the  triangle  ABC,  divide  the  op- 
posite side  B  C,  so  that  B  D  :  D  C  :  : 
A  B  :  A  C  ;    then  will  the  line  A  D 
bisect  the  angle  BAG. 

Through  the  point  G  draw  G  E  parallel  to  AD,  meeting 
B  A  produced  in  E.  Then,  by  hypothesis,  B  D  :  D  G  :  : 
A  B  :  A  G  ;  and  since  AD  is  parallel  to  E  G,  B  D  :  D  G  :  : 
A  B  :  A  E  (Prop.  XVII.)  ;  then  AB:AG::AB:AE 
(Prop.  X.  Bk.  II.)  ;  consequently  A  G  is  equal  to  A  E  ; 
hence  the  angle  A  E  G  is  equal  to  the  angle  AGE  (Prop. 
VII.  Bk.  L).  But,  since  G  E  and  AD  are  parallels,  the 
angle  AE  G  is  equal  to  the  opposite  exterior  angle  BAD, 
and  the  angle  A  G  E  is  equal  to  the  alternate  angle  DAG 
(Prop.  XXTL  Bk.  I.)  ;  hence  tlie  angles  BAD,  DAG 
are  equal,  and  consequently  the  straight  line  A  D  bisects 
the  angle  BAG. 

Proposition  XXI. — Theorem. 

258.  If  the  exterior  angle  formed  by  producing  one  of 
the  sides  of  any  triangle  be  bisected  by  a  straight  line 
which  meets  the  base  produced,  the  distances  from  the  ex- 
tremities of  the  base  to  the  point  where  the  bisecting  line 
meets  the  base  produced,  will  be  to  each  other  as  the  other 
two  sides  of  the  triangle. 

Let  the  exterior  angle  G  AE,  formed  by  producing  the 
side  B  A  of  the  tria  igle  A  B  G,  be  bisected  1)y  the  straight 


BOOK   IV. 


99 


line  AD,  which  meets  the  side 
B  C  produced  in  D,  then  will 
BD  :  DC  :  :  AB:  AC. 
Through  C  draw  CF  parallel 
to  A  D  ;  then  the  angle  A  C  F  is 
equal  to  the  alternate  angle  CAD, 
and  the  exterior  angle  D  A  E  is 
equal  to  the  interior  and  opposite  angle  C  F  A  (Prop. 
XXII.  Bk.  I.).  But,  by  hypothesis,  the  angles  CAD, 
DAE  are  equal ;  consequently  the  angle  A  C  F  is  equal 
to  the  angle  C  F  A  ;  hence  the  triangle  A  C  F  is  isosceles, 
and  the  side  AC  is  equal  to  the  side  AF  (Prop.  VIII. 
Bk.  I.).  Again,  since  A  D  is  parallel  to  F  C,  B D  :  D  C  :  : 
BA:  AF  (Prop.  XVII.  Cor.  1),  and  substituting  AC 
in  the  place  of  its  equal  A  F,  we  have 

BD:DC::BA:AC. 


Proposition  XXII.  —  Theorem. 

259.  Equiangular  triangles  have  their  homologous  sides 
proportional^  and  are  similar. 

Let  the  two  triangles  A  B  C,  D  C  E  ^ 

be   equiangular;   the   angle   BAC  /    \ 

being  equal  to  the  angle  CDE,  the 
angle  A  B  C  to  the  angle  D  C  E,  and 
the  angle  A C B  to  the  angle  DEC, 
then  the  homologous  sides  will  be 
proportional,  and  we  shall  have 

BC:CE::AB:CD::AC:DE. 

For,  let  the  two  triangles  be  placed  so  that  two  homol- 
ogous sides,  B  C,  C  E,  may  join  each  other,  and  be  in  the 
same  straight  line  ;  and  produce  the  sides  B  A,  E  D  till 
they  meet  in  F. 

Since  B  C  E  is  a  straight  line,  and  the  angle  B  C  A  is 
equal  to  the  angle  C  E  D,  AC  is  parallel  to  F  E  (Prop. 
XXI.  Bk.  I.)  ;  also,  since  the  angle  A  B  C  is  equal  to  the 


100  ELEMENTS  OF  GEOMETRY. 

angle  D  C  E,  the  line  B  F  is  parallel 
to  the  line  CD.  Hence  the  figure 
A  C  D  F  is  a  parallelogram  ;  and, 
consequently,  A  F  is  equal  to  CD, 
and  AC  to  FD  (Prop.  XXXI. 
Bk.  I.).  B  C  E 

In  the  triangle  B  E  F,  since  the  line  A  C  is  parallel  to 
the  side  F  E,  we  have  BC  :  CE  :  :  BA  :  AF  (Prop. 
XYII.)  ;  or,  substituting  CD  for  its  equal,  AF, 

BC:CE::BA:CD. 

Again,  C  D  is  parallel  to  B  F ;  therefore,  B  C  :  C  E  :  : 
F  D  :  D  E  ;  or,  substituting  A  C  for  its  equal  F  D, 

B  C  :  C  E  :  :  A  C  :  D  E. 

And,  since  both  tliese  proportions  contain  the  same  ratio 
B  C  :  C  E,  we  have  (Prop.  X.  Bk.  II.) 

AC  :  DE  :  :  BA:  CD. 
Hence,  the  equiangular  triangles  B  A  C,   C  D  E  have 
their   homologous   sides   proportional ;  and  consequently 
the  two  triangles  are  similar  (Art.  210). 

260.  Cor.  Two  triangles  having  two  angles  of  the  one 
equal  to  two  angles  of  the  other,  each  to  each,  are  similar; 
since  the  third  angles  will  also  be  equal,  and  the  two  tri- 
angles be  equiangular. 

261.  Scholium.  In  similar  triangles,  the  homologous 
sides  are  opposite  to  the  equal  angles  ;  thus  the  angle 
A  C  B  being  equal  to  DEC,  the  side  A  B  is  homologous 
to  D  C  ;  in  like  manner,  A  C  and  D  E  are  homologous. 

Proposition  XXIII.  —  Theorem. 

262.  Triangles  lohich  have  their  homologous  sides  pro- 
portional,  are  equiang"ular  and  similar. 

Let  the  two  triangles  ABC,  D  E  F  have  their  sides  pro- 
portional, so  that  we  have  BC :  EF  : :  AB  :  DE  ; :  AC  :  D  F; 


«  BOOK   IV.  101 

tlien  will  the  triangles 
have  their  angles  equal ; 
namely,  the  angle  A 
equal  to  the  angle  D, 
the  angle  B  to  the  angle 
E,  and  the  angle  C  to 
the  angle  F. 

At  the  point  E,  in  the    B  C         G 

straight  line  EF,  make  the  angle  FEG  equal  to  the  angle 
B,  and  at  the  point  F,  the  angle  E  F  G  equal  the  angle  C ; 
the  third  angle  G  will  be  equal  to  the  third  angle  A 
(Prop.  XXVIII.  Cor.  2,  Bk.  I.)  ;  and  the  two  triangles 
ABC,  E  F  G  will  be  equiangular.  Therefore,  by  the 
last  theorem,  we  have 

B  C  :  E  F  :  :  A  B  :  E  G ; 

but,  by  hypothesis,  we  have 

BC  :  EF:  :  AB  :  DEj    ^       »..„.,...... 

hence,  E  G  is  equal  to  D  E. 

By  the  same  theorem,  we  also  have 

BC:EF:  :  AC  :FG; 

and,  by  hypothesis, 

BC:EF::AC:DF; 

hence  F  G  is  equal  to  D  F.  Hence,  the  triangles  E  G  F, 
D  E  F,  having  their  three  sides  equal,  each  to  each,  are 
themselves  equal  (Prop.  XVIII.  Bk.  I.).  But,  by  con- 
struction, the  triangle  E  G  F  is  equiangular  with  the  tri- 
angle ABC;  hence  the  triangles  D  E  F,  ABC  are  also 
equiangular  and  similar. 

263.  Scholium.  The  two  preceding  propositions,  togctlier 
with  that  relating  to  the  square  of  the  hypothenuse  (Art. 
237),  are  the  most  important  and  fertile  in  results  of  any 
in  Geometry.  They  are  almost  sufficient  of  themselves 
for  all  applications  to  subsequent  reasoning,  and  for  the 
9* 


102 


ELEMENTS    OF   GEQ^IETRY. 


solution  of  all  problems  ;  since  the  general  properties  of 
triangles  include,  by  implication,  those  of  all  figures. 

Proposition    XXIY.  —  Theorem. 

264.  Tvw  triangles,  which  have  an  angle  of  the  one 
equal  to  an  angle  of  the  other,  and  the  sides  containing 
these  angles  proportional,  are  similar. 

Let  the  two  triangles  ABC, 
I)  E  F  have  the  angle  A  equal  to 
the  angle  D,  and  the  sides  contain- 
ing these  angles  proportional,  so 
that  AB  :  DE  :  :  AC  :  DF;  then 
tlie  triangles  are  similar. 

Take  AG  equal  D  E,  and  draw 
GH  parallel  to  BC.     The  angle  AGH  will  b^  equal  to 
fhe  angle' A 'H  (3  (Prop.  XXII.  Bk.  I.);   and  the  triangles 
AQ.H,  A.BC  will  "be  equiangular;  hence  we  shall  have 
<'i>"U.'^  ':^  .c      AB:  AG:  :  AC  :  AH. 
But,  by  hypothesis, 

A  B  :  D  E  :  :  A  C  :  D  F ; 

and,  by  construction,  AG  is  equal  to  D  E  ;  hence  AH  is 
equal  to  D  F.  Therefore  the  two  triangles  A  G  H,  D  E  F, 
having  two  sides  and  the  included  angle  of  the  one  equal 
to  two  sides  and  the  included  angle  of  the  other,  each  to 
each,  are  themselves  equal  (Prop.  Y.  Bk.  I.).  But  the 
triangle  A  G  H  is  similar  to  ABC;  therefore  D  E  F  is 
also  similar  to  ABC. 


Proposition  XXY . : —  Theorem. 

265.  Tivo  triangles,  which  have  their  sides,  taken  two 
and  two,  either  parallel  or  perpendicular  to  each  other, 
are  similar. 

Let  the  two  triangles  ABC,  D  E  F  have  the  side  A  B 
parallel  to  the  side  D  E,  B  C  parallel  to  E  F,  and  A  C 


BOOK   IV.  103 

parallel   to   D  F  ;  these  triangles  will 
then  be  similar. 

For,  since  the  side  AB  is  parallel  to 
the  side  D  E,  and  B  C  to  E  F,  the  angle 
ABC  is  equal  to  the  angle  DEF 
(Prop.  XXYI.  Bk.  I.).  Also,  since 
AC  is  parallel  to  DF,  the  angle  ACB 
is  equal  to  the  angle  D FE,  and  the  angle  BAC  to  EDF; 
therefore  the  triangles  ABC,  DEF  are  equiangular; 
hence  they  are  similar  (Prop.  XXII.) . 

Again,  let  the  two  triangles  ^ 

ABC,  DEF  have  the  side 
D  E  perpendicular  to  the  side 
AB,  DF  perpendicular  to  AC, 
and  E  F  perpendicular  to  B  C ; 
these  triangles  are  similar. 

Produce  F  D  till  it  meets  A  C 
at  Gr ;  tlien  the  angles  D  G  A,  D  E  A  of  the  quadrilateral 
A  E  D  G  are  two  right  angles  ;  and  since  all  the  four  an- 
gles are  together  equal  to  four  right  angles  (Prop.  XXIX. 
Cor.  1,  Bk.  I.),  the  remaimng  two  angles,  E  D  G,  E  A  G, 
are  together  equal  to  two  right  angles.  But  the  two 
angles  E  D  G,  EDF  are  also  together  equal  to  two  right 
angles  (Prop.  I.  Bk.  I.)  ;  hence  the  angle  E  D  F  is  equal 
to  E AG  or  BAC. 

The  two  angles,  G  F  C,  G  C  F,  in  the  right-angled  trian- 
gle F  G  C,  are  together  equal  to  a  right  angle  (Prop. 
XXYIII.  Cor.  5,  Bk.  I.),  and  the  two  angles  GFC,  GFE 
are  together  equal  to  the  right  angle  E  F  C  (Art.  34,  Ax. 
9)  ;  therefore  G  F  E  is  equal  to  G  C  F,  or  D  F  E  to  B  C  A. 
Therefore  the  triangles  ABC,  DEF  have  two  angles  of 
the  one  equal  to  two  angles  of  the  other,  each  to  each ; 
hence  they  are  similar  (Prop.  XXII.  Cor.). 

266.  Scholium.  When  the  two  triangles  have  their  sides 
parallel,  the  parallel  sides  are  homologous  ;  and  when- 
they  have  them  perpendicular,  the  perpendicular  sides  are 


104  ELEMENTS  OP  GEOMETRY. 

homologous.     Thus,  D  E  is  homologous  with  A  B,  D  F 
with  A  0,  aiidEF  with  B  C. 

Proposition   XX YI.  —  Theorem. 

267.  In  any  triangle,  if  a  line  he  drawn  parallel  to  the 
base,  all  lines  drawn  from  the  vertex  icill  divide  the  par- 
allel and  the  base  proportionally. 

In  the  triangle  BAG,  let  DE  a 

be  drawn  parallel  to  the  base  B  C  ; 
then  will  the  lines  A  F,  A  G,  A  H, 
drawn  from  the  vertex,  divide  the 
parallel  D  E,  and  the  base  B  C,  so 
that 
DI:BF::IK:FG::KL:GH. 

For,  since  D I  is  parallel  to  B  F,  the  triangles  AD  I 
and  A  B  F  are  equiangular  ;  and  we  have  (Prop.  XXII.), 

DI:  BF:  :  AI:  AF; 

and  since  I K  is  parallel  to  F  G,  we  have  in  like  manner, 

AI:AF::IK:FG; 

and,  since  these  two  propositions  contain  the  same  ratio, 
A I  :  AF,  we  shall  have  (Prop.  X.  Cor.  1,  Bk.  II.), 

DI:BF::IK:FG. 

In  the  same  manner,  it  may  be  shown  that 

IK:FG:  :KL:  GH:  :LE:HC. 

Therefore  the  line  D  E  is  divided  at  the  points  I,  K,  L,  as 
the  base  B  C  is,  at  the  points  F,  G,  H. 

268.  Cor.  If  B  C  were  divided  into  equal  parts  at  the 
points  F,  G,  H,  the  parallel  D  E  would  also  be  divided 
into  equal  parts  at  the  points  I,  K,  L. 

Proposition  XXVII.  —  Theorem. 

269.  In  a  right-angled  triangle,  if  a  perpendicular  is 
drawn  from  the  vertex  of  the  right  angle  to  the  hypothec 


BOOK   IV.  f^  105 

nuse,  the  triangle  icill  be  divided  into  ttvo  triangles  simi- 
lar to  the  given  triangle  and  to  each  other. 

In  the  right-angled  triangle  ABC, 
from  the  vertex  of  the  right  angle 
B  A  C,  let  AD  be  drawn  perpendicu- 
lar to  the  hypothenuse  B  C ;  then  the 
triangles  BAD,  D  A  C  will  be  simi- 
lar to  the  triangle  ABC,  and  to  each 
other. 

For  the  triangles  BAD,  B  A  C  have  the  common  angle 
B,  the  right  angle  B  D  A  equal  to  the  right  angle  B  xV  C, 
and  therefore  the  third  angle,  B  AD,  of  the  one,  equal  to 
the  third  angle,  C,  of  the  other  (Prop.  XXVIII.  Cor.  2, 
Bk.  I.)  ;  hence  these  two  triangles  are  equiangular,  and 
consequently  are  similar  (Prop.  XXII.).  In  the  same 
manner  it  may  be  shown  that  the  triangles  D  A  C  and 
B  A  C  are  equiangular  and  similar.  The  triangles  BAD 
and  D  A  C,  being  each  similar  to  the  triangle  B  A  C,  are 
similar  to  each  other. 

270.  Cor,  1.  Each  of  the  sides  containing  the  right 
angle  is  a  mean  proportional  between  the  hypothenuse 
and  the  part  of  it  which  is  cut  off  adjacent  to  that  side 
by  the  perpendicular  from  the  vertex  of  the  right  angle. 

For,  the  triangles  BAD,  BAC  being  similar,  their 
homologous  sides  are  proportional ;  hence 

BD  :  BA:  :BA:  BC; 

and,  the  triangles  D  A  C,  B  A  C  being  also  similar, 

DC:AC::AC:BC; 

hence  each  of  the  sides  A  B,  A  C  is  a  mean  proportional 
between  the  hypothenuse  and  the  part  cut  off  adjacent 
to  that  side. 

271.  Cor.  2.  The  perpendicular  from  the  vertex  of  the 
right  angle  to  the  hypothenuse  is  a  mean  proportional  be- 
tween the  two  parts  into  which  it  divides  the  hypothenuse. 


106  ELEMENTS  OF  GEOMETRY. 

For,  since  the  triangles  ABD,  ADC  are  similar,  by 
comparing  their  homologous  sides  we  have 

BD:AD::AD:DC; 

hence,  the  perpendicular  A  D  is  a  mean  proportional  be- 
tween  the  parts  D  B,  D  C  into  which  it  divides  the  hy- 
pothenuse  B  C. 

Proposition  XXYIII.  —  Theorem. 

272.  Two  triangles,  having  an  angle  in  each  equal,  are 
to  each  other  as  the  rectangles  of  the  sides  which  contain 
the  equal  angles. 

A 

Let  the  two  triangles  ABC,  A  D  E 
have  the  angle  A  in  common  ;  then 
will  the  triangle  ABC  be  to  the  tri- 
angle ADEasABxAC  toADX 
AE.  

Join  B  E  ;  then  the  triangles  A  B  E,     B  C 

A  D  E,  having  the  common  vertex  E,  and  their  bases  in 
the  same  line,  AB,  have  the  same  altitude,  and  are  to 
each  other  as  their  bases  (Prop.  VI.  Cor.)  ;  hence 

ABE:ADE::AB:AD. 

In  like  manner,  since  the  triangles  ABC,  ABE  have 
the  common  vertex  B,  and  their  bases  in  the  same  line, 
A  C,  we  have 

ABC:ABE::AC:AE. 

By  multiplying  together  the  corresponding  terms  of 
these  proportions,  and  omitting  the  common  term  ABE, 
we  have  (Prop.  XIII.  Bk.  II.), 

ABC:ADE::ABxAC:ADxAE. 

273.  Cor.  If  the  rectangles  of  the  sides  containing  the 
equal  angles  were  equivalent,  the  triangles  would  be 
equivalent. 


BOOK   17. 


'% 


107 


C    E  F 

proportional    (Art. 


Proposition  *XXIX.  —  Theorem. 

274.  Similar  triang-les  are  to  each  other  as  the  squares 
described  on  their  homologous  sides. 

Let  A  B  C,  D  E  F  be  two  similar  A 
triangles,  and  let  AC,  D  F  be  ho- 
mologous sides ;  then  the  triangle 
ABC  will  be  to  the  triangle  D  E  F 
as  the  square  on  AC  is  to  the  square 
on  D  F. 

For,  the  triangles  being  similar, 
they   have   their   homologous    sides 
210);  therefore 

AB:  DE  :  :  AC:  DF; 

and  multiplying  the  terms  of  this  proportion  by  the  cor- 
responding terms  of  the  identical  proportion, 

AC:DF:  :  AC:DF, 
we  have  (Prop.  XIII.  Bk.  II.), 

ABXAC:DEXDF::  AC^  :  DF^ 

But,  by  reason  of  the  equal  angles  A  and  D,  the  triangle 
ABC  is  to  the  triangle  DEFasAB  X  AC  is  toDExDF 
(Prop.  XXVIII.)  ;  consequently  (Prop.  X.  Bk.  II.), 

A  B  C  :  D  E  F  :  :  AC^  :  DFI 

Therefore,  the  two  similar  triangles  ABC,  D  E  F  are 
to  each  other  as  the  squares  described  on  the  homologous 
sides  AC,  D  F,  or  as  the  squares  described  on  any  other 
two  homologous  sides. 


Proposition  XXX.  —  Theorem. 

275.  Similar  polygons  may  be  divided  into  the  same 
number  of  triangles  similar  each  to  each,  and  similarly 
situated. 


108 


Elements  of  geometry. 


Let  ABODE, 
FGHIK  be  two 
similar  polygons ; 
they  may  be  divid- 
ed into  the  same 
number  of  trian- 
gles similar  each 
to  each,  and  similarly  situated.  From  the  homologous  an- 
gles A  and  F,  draw  the  diagonals  A  C,  A  D  and  F  H,  F  I. 

The  two  polygons  being  similar,  the  angles  B  and  G, 
which  are  homologous,  must  be  equal,  and  the  sides  AB, 
B  C  must  also  be  proportional  to  F  G,  G  H  (Art.  210)  ; 
that  is,  A  B  :  F  G  :  :  B  C  :  G  H.  Therefore  the  triangles 
A  B  C,  F  GH  have  an  angle  of  the  one  equal  to  the  angle 
of  the  other,  and  the  sides  containing  these  angles  propor- 
tional ;  hence  they  are  similar  (Prop.  XXIV.)  ;  conse- 
quently the  angle  B  C  A  is  equal  to  the  angle  G  H  F. 
These  equal  angles  being  taken  from  the  equal  angles 
BCD,  G  H  I,  the  remaining  angles  A  C  D,  F  H  I  will  be 
equal  (Art.  34,  Ax.  3).  But,  since  the  triangles  ABC, 
F  G  H  are  similar,  we  have 

AC:FH:  :  BC:  GH; 

and,  since  the  polygons  are  similar  (Art.  210), 

BC:GH::CD:HI; 
hence  (Prop.  X.  Cor.  1,  Bk.  II.), 

AC:FH::CD:HI. 

But  the  terms  of  the  last  proportion  are  the  sides  about 
the  equal  angles  A  C  D,  F  H I ;  hence  the  triangles  A  C  I), 
FHI  are  similar  (Prop.  XXIY.).  In  the  same  mannjer, 
it  may  be  shown  that  the  corresponding  triangles  A  D  E, 
F I K  are  similar ;  hence  the  similar  polygons  may  be 
divided  into  the  same  number  of  triangles  similar  each  to 
each,  and  similarly  situated. 

276.    Cor.    Conversely^  if  tiuo  polygons  are  composed 


BOOK  IV. 


109 


of  the  same  numher  of  similar  triangles^  and  similarly 
situated,  the  tvw  polygons  are  similar. 

For  tlie  similarity  of  the  corresponding  triangles  give 
the  angles  ABC  equal  to  F  G  H,  B  C  A  equal  to  G  H  F, 
and  A  C  D  equal  to  F  H I ;  hence,  BCD  equal  to  G  H  I, 
likewise  C  D  E  equal  to  H  1 K,  <fec.     Moreover,  we  have 

AB:FG:  :BC:  GH:  :  AC:FH::CD:HI,&c.; 

therefore  the  two  polygons  have  their  angh 
their  sides  proportional ;  hence  they  are  sii 

Proposition  XXXI.  —  Theor 

277.    The  perimeters  of  similar  polygons  are^ 
other  as  their  homologous  sides ;  and  their  areas  art  to 
each  other  as  the  squares  described  on  these  sides. 

Let  ABCDE, 
FGHIK  be  two 
similar  polygons ; 
then  their  perim- 
eters are  to  each 
other  as  their  ho- 
mologous sides 
A  B  and  F  G,  B  C  and  GH,  <S:c 
each  other  as  AB^  is  to  F~G^  BC^  to  GH^  <fec. 

First,  Since  the  two  polygons  are  similar,  we  have 
AB:FG:  :BC  :  GH:  :  CD  :HI,  cfcc. 

Now  the  sum  of  the  antecedents  A  B,  B  C,  CD,  <fec., 
which  compose  the  perimeter  of  the  first  polygon,  is  to 
the  sum  of  the  consequents  F  G,  GH,  HI,  <fec.,  which 
compose  the  perimeter  of  the  second  polygon,  as  any  one 
antecedent  is  to  its  consequent  (Prop.  XI.  Bk.  II.)  ;  there- 
fore, as  any  two  homologous  sides  are  to  each  other,  or  as 
A  B  is  to  F  G. 

Secondly.  From  the  homologous  angles  A  and  F,  draw 


and  their  areas  are  to 

2 


110  ELEMENTS   OF   GEOMETRY. 

the  diagonals  A  C,  A  D  and  F  H,  F  I.  Then,  smce  the  tri- 
angles ABC,  F  GH  are  similar,  the  triangle 

A B  C  :  F  G H  :  :  AC^  :  FH^ 

(Prop.  XXIX.)  ;  and,  since  the  triangles  AC  D,  F HI  are 

similar,  the  triangle  A  C  D  :  F  H I  :  :  AC^  :  FhI     But 

the  ratio  AC  :  FH  is  common  to  both  of  the  propor- 
tions; therefore  (Prop.  X.  Bk.  II.), 

ABC:FGH::ACD:FHI. 

By  the  same  mode  of  reasoning,  it  may  be  proved  that 
ACD  :FHI:  :  ADE:FIK, 

and  so  on,  if  there  were  more  triangles.  Therefore  the 
sum  of  the  antecedents  ABC,  ACD,  ADE,  which  com- 
pose the  area  of  the  polygon  A  B  C  D  E,  is  to  the  sum  of 
the  consequents  FGH,  FHI,  FIK,  which  compose  the 
area  of  the  polygon  F  G  H I K,  as  any  one  antecedent 
AB  C  is  to  its  consequent  FGH  (Prop.  XI.  Bk.  II.),  or 
as  A  B"  is  to  F  G  ;  hence  the  areas  of  similar  polygons 
are  to  each  other  as  the  squares  described  on  their  ho- 
mologous sides. 

278.  Cor.  1.  The  perimeters  of  similar  polygons  are 
also  to  each  other  as  their  corresponding  diagonals. 

279.  Cor.  2.  The  areas  of  similar  polygons  are  to  each 
other  as  the  squares  described  on  their  corresponding 
diagonals. 

Proposition  XXXII. — Theorem. 

280.  A  chord  in  a  circle  is  a  mean  proportional  between 
the  diameter  and  the  part  of  the  diameter  cut  off  between 
one  extremity  of  the  chord  and  a  perpendicular  drawn 
from  the  other  extremity  to  the  diameter. 

Let  A  B  be  a  chord  in  a  circle,  B  C  a  diameter  drawn 
from   one   extremity   of  A  B,   and   A  D  a  perpendicular 


BOOK   17.  Ill 

drawn  from  the  other  extremity  to 
BC;  then 

BD:AB::AB:BC. 

Join  A  C  ;  then  the  triangle 
ABC,  described  in  a  semicircle,  is 
right-angled  at  A  (Prop.  XVIII. 
Cor.  2,  Bk.  III.)  ;  and  the  triangle 
B  A  D  is  similar  to  the  triangle  B  A  C  (Prop.  XXVIl.)  ; 
hence,  we  have  (Prop.  XXVII.  Cor.  1), 
BD  :  AB:  :  AB:BC; 
therefore  the  chord  A  B  is  a  mean  proportional  between 
the  diameter  B  C,  and  the  part,  B  D,  cut  off  between  the 
extremity  of  the  chord  and  the  perpendicular  from  the 
other  extremity. 

281.  Cor.  If  from  any  point.  A,  in  the  circumference  of 
a  circle,  a  perpendicular,  AD,  be  drawn  to  the  diameter 
BC,  the  perpendicular  will  be  a  mean  proportional  be- 
tween the  parts  BD,  D  C  into  which  it  divides  the  diam- 
eter. 

For,  joining  AB  and  AC,  we  have  the  triangle  ABC, 
right-angled  at  xV,  and  the  triangles  BAD,  D  A  C  similar 
to  it  and  to  each  other  (Prop.  XXYII.) ;  therefore 
(Prop.  XXYII.  Cor.  2), 

BD  :  AD  :  :  AD  :  DC, 

or,  what  amounts  to  the  same  thing  (Prop.  III.  Bk.  II.), 

B  D  X  D  C  =  ADI 

Scholium.  A  part  of  a  straight  line  cut  off  by  another 
is  called  a  segment  of  the  line.  Thus  BD,  D  C  are  seg- 
ments of  the  diameter  B  C. 

Proposition  XXXIII.  —  Theorem. 

282.  If  tivo  chords  in  a  circle  intersect  each  other ^  the 
segments  of  the  one  are  reciprocally  proportional  to  the 
segments  of  the  other. 


112 


ELEMENTS    OF   GEOMETRY. 


Let  A B,  CD  be  two  chords,  which 
intersect  each  other  at  E  ;  then  will 
AE:DE::EC:  EB. 

Join  A  C  and  B  D.  In  the  triangles 
A  EC,  BED,  the  angles  at  E  are 
equal  being  vertical  angles  (Prop. 
IV.  Bk.  I.)  ;  the  angle  A  is  equal  to 
the  angle  D,  being  measured  by  half  the  same  arc,  BC 
(Prop.  XVIII.  Cor.  1,  Bk.  III.)  ;  for  the  same  reason,  the 
angle  C  is  equal  to  the  angle  B ;  the  triangles  are  there- 
fore similar  (Prop.  XXII.),  and  their  homologous  sides 
give  the  proportion, 

AE:DE::EC:EB. 

283.  Cor.  Hence,  AExEB  =  DExEC;  there- 
fore the  rectangle  of  the  two  segments  of  the  one  chord  is 
equal  to  the  rectangle  of  the  two  segments  of  the  other. 

Proposition   XXXIV .  —  Theorem  . 

284.  If  from  the  same  point  without  a  circle  tioo  secants 
he  dravm^  terminating  in  the  concave  arc^  the  whole  se- 
cants will  be  reciprocally  proportional  to  their  external 

sesrments. 

p. 
Let  E  B,  E  C  be  two  secants  drawn 

from  the  point  E  without  a  circle,  and 

terminating  in  the  concave  arc  at  the 

points  B  and  C  ;  then  will 

EB:EC::ED:EA. 

For,  joining  AC,  BD,  the  triangles 
A  E  C,  BED  have  the  angle  E  com- 
mon ;  and  the  angles  B  and  C,  being  ^ 
measured  by  lialf  the  same  arc,  A  D,  are  equal  (Prop. 
XVIII.  Cor.  1,  Bk.  III.)  ;  these  triangles  are  therefore 
similar  (Prop.  XXII.  Cor.),  and  their  homologous  sides 
give  the  proportion, 

EB:  EC  :  :  ED  :EA. 


BOOK   IV. 

/ 


113 


285.  Cor,  Hence,  EBxEA  =  ECxED;  therefore 
the  rectangle  contained  bj  the  whole  of  one  secant  and  its 
external  segment  is  equivalent  to  the  rectangle  contained 
by  the  whole  of  the  other  secant  and  its  external  segment. 

Proposition  XXXY .  —  Theorem. 

286.  If  from  a  point  without  a  circle  there  he  draicn  a 
tangent  terminating-  in  the  circumference^  and  a  secant 
terminating  in  the  concave  arc^  the  tafigent  ivill  be  a  mean 
proportional  between  the  whole  secant  and  its  external 
segment. 

From  the  point  E  let  the  tangent 
E  A,  and  the  secant  E  C,  be  drawn ; 
then  will  EC:EA::EA:ED. 

For,  joining  A  D  and  A  C,  the 
triangles  E  A  D,  E  A  C  have  the 
angle  E  common  ;  also,  the  angle 
E  A  D  formed  by  a  tangent  and  a 
chord  has  for  its  measure  half  the 
arc  AD  (Prop.  XX.  Bk.  III.),  and  the  angle  C  has  the 
same  measure  ;  therefore  the  angle  E  A  D  is  equal  to  the 
angle  C ;  hence  the  two  triangles  are  similar  (Prop.  XXII. 
Cor.),  and  give  the  proportion, 

EC:EA::EA:ED. 

287.  Cor.  Hence,  EA^  =  E  C  X  E  D  ;  therefore  the 
square  of  the  tangent  is  equivalent  to  the  rectangle  con- 
tained by  the  whole  secant  and  its  external  segment. 


Proposition  XXXY  I.  —  Theorem. 

288.  If  any  angle  of  a  triangle  is  bisected  by  a  line 
terminating  in  the  opposite  side,  the  rectangle  of  the  other 
tvjo  sides  is  equivalent  to  the  square  of  the  bisecting  line 
plus  the  rectangle  of  the  segments  of  the  third  side. 

10* 


114 


ELEMENTS   OF   GEOMETRY. 


Let  the  triangle  ABC  have  the 
angle  BAG  bisected  by  the  straight 
line  AD  terminating  in  the  oppo- 
site side  B  C  ;  then  the  rectangle 
B  A  X  AC  is  equivalent  to  the 
square  of  AD  plus  the  rectangle 
B  D  X  DC.  Describe  a  circle 
througli  the  three  points  A,  B,  C  ; 
produce  A  D  till  A  meets  the  circumference  at  E,  and  join 
CE. 

The  triangles  BAD,  EAC  have,  by  hypothesis,  the 
angle  BAD  equal  to  the  angle  EAC;  also  the  angle  B 
equal  to  the  angle  E,  being  measured  by  half  of  the  same 
arc  AC  (Prop.  XVIII.  Cor.  1,  Bk.  III.)  ;  these  triangles 
are  therefore  similar  (Prop.  XXII.  Cor.),  and  their  ho- 
mologous sides  give  the  proportion, 

B  A  :  A  E  :  :  A  D  :  A  C  ; 

hence,  BAxAC  =  AExAD. 

But  A  E  is  equal  to  A  D  +  D  E,  and  multiplying  each 
of  these  equals  by  AD,  we  have, 

A  E  X  A  D  =  AD'  +  A  D  X  D  E  ; 

now,  AD   X  D  E   is   equivalent  to  B  D  X  DC    (Prop. 
XXXIII.  Cor.)  ;  hence 

B  A  X  A  C  =  ad'  +  B  D  X  D  C. 


Proposition  XXXVII.  —  Theorem. 

289.  The  rectangle  contained  by  any  two  sides  of  a 
triangle  is  equivalent  to  the  rectangle  contained  by  the 
diameter  of  the  circumscribed  circle  and  the  perpendicular 
drawn  to  the  third  side  from  the  vertex  of  the  opposite 
angle. 

In  any  triangle  A  B  C,  let  A  D  be  drawn  perpendicular 
to  B  C ;  and  let  E  C  be  the  diameter  of  the  circle  circum- 


BOOK   IV. 


115 


scribed  about  the  triangle  ;  then 
will  A  B  X  A  C  be  equivalent  to 
A  D  X  0  E. 

For,  joining  AE,  the  angle  E  AC 
is  a  right  angle,  being  inscribed  in 
a  semicircle  (Prop.  XVIII.  Cor.  2, 
Bk.  III.)  ;  and  the  angles  B  and 
E  are  equal,  being  measured  by  half  of  the  same  arc,  A  C 
(Prop.  XVIII.  Cor.  1,  Bk.  III.) ;  hence  the  two  right- 
angled  triangles  are  similar  (Prop.  XXII.  Cor.),  and  give 
the  proportion  AB:CE::AD:AC;  hence 
ABXAC  =  CEXAD. 

290.   Cor.   If  these  equals  be  multiplied  by  BC,  we 
shall  have 

ABXACXBC  =  CEXADXBC. 

But  AD  X  B  C  is  double  the  area  of  the  triangle 
(Prop.  YI.)  ;  therefore  the  product  of  the  three  sides  of 
a  triangle  is  equal  to  its  area  multiplied  by  twice  the 
diameter  of  the  circumscribed  circle. 


Proposition  XXXVIII.  —  Theorem. 

291.  Tlie  rectangle  contained  by  the  diagonals  of  a 
quadrilateral  inscribed  in  a  circle  is  equivalent  to  the  sum 
of  the  two  rectangles  of  the  opposite  sides. 

Let  A  B  C  D  be  any  quadrilat- 
eral inscribed  in  a  circle,  and  A  C, 
B  D  its  diagonals  ;  then  the  rec- 
tangle A  C  X  B  D  is  equivalent  to 
the  sum  of  the  two  rectangles 
AB  X  CD,  AD  X  BC. 

For,  draw  BE,  making  the  angle 
ABE  equal  to  the  angle  C  B  D  ; 
to  each  of  these  equals  add  the  angle  E  BD,  and  we  shall 
have  the  angle  A  B  D  equal  to  the  angle  E  B  C  ;  and  th© 


IIG 


ELEMENTS  OF    GEOMETRY. 


angle  ADB  is  equal  to  the  angle 
BCE,  being  in  the  same  segment 
(Prop.  XVm.  Cor.  1,  Bk.  III.)  ; 
therefore  the  triangles  ABD,  B  C  E 
are  similar  ;  hence  the  proportion, 

AD  :  BD  :  :  CE  :  BC; 
and,  consequently, 

ADXBC  =  BDXCE. 
Again,  since  the   angle  A  B  E  is  equal  to  the  angle 
C  B  D,  and  the  angle  B  A  E  is  equal  to  the  angle  B  D  0, 
being  in  the  same  segment  (Prop.  XVIII.  Cor.  1,  Bk.  HI.)? 
the  triangles  A  B  E,  B  C  D  are  similar  ;  hence, 

AB:AE::BD:CD; 
and  consequently, 

ABxCD  =  AExBD. 
By  adding  the  corresponding  terms  of  the  two  equations 
obtained,  and  observing  that 

BDxAE  +  BDxCE  =  BD(AE  +  CE)  =  BDxAC, 
we  have 

BD  X  AC  =  AB  X  CD  + AD  X  BC. 


E 


Proposition  XXXIX. — Theorem. 

292.    The  diagonal  of  a  square  is  incommensurable  with 
its  side. 

Let  A  B  C  D  be  any  square,  and 
A  C  its  diagonal ;  then  A  C  is  in- 
commensurable with  the  side  A  B. 

To  find  a  common  measure,  if 
there  be  one,  we  must  apply  A  B, 
or  its  equal  CB,  to  C  A,  as  often  as 
it  can  be  done.  In  order  to  do  this, 
from  the  point  C  as  a  centre,  with 
a  radius  C  B,  describe  the  semicircle  F  B  E,  and  produce 
AC  to  E.    It  is  evident  that  CB  is  contained  once  in  AC, 


y 


BOOK   IV.  117 

with  a  remainder  AF,  which  remainder  must  be  com- 
pared with  B  0,  or  its  equal,  A  B. 

The  angle  ABC  being  a  right  angle,  A  B  is  a  tangent 
to  the  circumference,  and  A  E  is  a  secant  drawn  from  the 
same  point,  so  that  (Prop.  XXXV.) 

A  F  :  A  B  :  :  A  B  :  A  E. 

Hence,  in  comparing  A  F  with  A  B,  the  equal  ratio  of 
A  B  to  A  E  may  be  substituted  ;  but  A  B  or  its  equal  C  F 
is  contained  twice  in  A  E,  with  a  remainder  A  F  ;  which 
remainder  must  again  be  compared  with  A  B. 

Thus,  the  operation  again  consists  in  comparing  AF 
with  A  B,  and  may  be  reduced  in  the  same  manner  to  the 
comparison  of  A  B,  or  its  equal  C  F,  with  A  E  ;  which 
will  result,  as  before,  in  leaving  a  remainder  A  F  ;  hence, 
it  is  evident  that  the  process  will  never  terminate  ;  conse- 
quently the  diagonal  of  a  square  is  incommensurable  with 
its  side. 

293.  Scholium,  The  impossibility  of  finding  numbers 
to  express  the  exact  ratio  of  the  diagonal  to  the  side  of  a 
square  has  now  been  proved  ;  but,  by  means  of  the  con- 
tinued fraction  which  is  equal  to  that  ratio,  an  approxi- 
mation may  be  made  to  it,  sufficiently  near  for  every 
practical  purpose. 


BOOK   V. 


PROBLEMS   RELATING  TO   THE   PRECEDING 
BOOKS. 

Problem  I. 

294.  To  bisect  a  given  straight  line,  or  to  divide  it  into 
two  equal  parts. 

Let  A  B  be  a  straight  line,  which  it  ;k'c 

is  required  to  bisect.  I 

From  the  point  A  as  a  centre,  with     j^ [E  ^ 

a  radius  greater  than  the  half  of  A  B,  I 

describe  an  arc  of  a  circle  ;  and  from  I 

the  point  B  as  a  centre,  with  the  same  ^^ 

radius,  describe  another  arc,  cutting  the  former  in  the 
points  C  and  D.  Through  C  and  D  draw  the  straight 
line  C  D  ;  it  will  bisect  A  B  in  the  point  E. 

For  the  two  points  C  and  D,  being  each  equally  distant 
from  the  extremities  A  and  B,  must  both  lie  in  the  per- 
pendicular raised  from  the  middle  point  of  AB  (Prop. 
XY.  Cor.,  Bk.  I.).  Therefore  the  line  CD  must  divide 
the  line  A  B  into  two  equal  parts  at  the  point  E. 

Problem  II. 

295.  From  a  given  point,  without  a  straight  line,  to 
draiv  a  perpendicular  to  that  line. 

Let  A  B  be  the  straight  line,  and  let  C  be  a  g'iven  point 
without  the  line. 


BOOK    V. 


119 


From  the  point  C  as  a  centre,  and 
with  a  radius  sufficiently  great,  describe 
an  arc  cutting  the  line  AB  in  two 
points,  A  and  B  ;  then,  from  the  points 

A  and  B   as  centres,    with    a   radius     ^':: 

greater  than  half  of  A  B,  describe  two 

arcs  cutting  each  other  in  D,  and  draw 

the  straight  line  C  D  ;  it  will  be  the  ^ 

perpendicular  required. 

For,  the  two  points  C  and  D  are  each  equally  distant 
from  the  points  A  and  B  ;  hence,  the  line  CD  is  a  perpen- 
dicular passing  through  the  middle  of  A  B  (Prop.  XV. 
Cor.,  Bk.  L). 


D 


B 


Problem  III. 

296.  At  a  given  point  in  a  straight  line  to  erect  a  per 
pendicular  to  that  line. 

Let  A  B  be  the  straight  line,  and  let  q 

D  be  a  given  point  in  it. 

In  the  straight  line  AB,  take  the 
points  A  and  B  at  equal  distances  from 
D ;  then  from  the  points  A  and  B  as  £- 
centres,  with  a  radius  greater  than  AD, 
describe  two  arcs  cutting  each  other  at  C  ;  through  C  and 
D  draw  the  straight  line  C  D  ;  it  will  be  the  perpendicular 
required. 

For  the  point  C,  being  equally  distant  from  A  and  B, 
must  be  in  a  line  perpendicular  to  the  middle  of  A  B 
(Prop.  XY.  Cor.,  Bk.  I.)  ;  hence  C  D  has  been  drawn 
perpendicular  to  A  B  at  the  point  D. 

297.  Scholium.  Tlie  same  construction  serves  for  mak- 
ing a  right  angle,  A  D  C,  at  a  given  point,  D,  on  a  given 
straight  line,  A  B. 


120 


ELEMENTS    OF   GEOMETRY. 


Problem  IY. 

298.    To  erect  a  perpendicular  at  the  end  of  a  g-ioen 
straight  line. 

Let  A  B  be  the  straight  line,  and 
B  the  end  of  it  at  which  a  perpen- 
dicular is  to  be  erected. 

From  any  point,  D,  taken  without 
the  line  A  B,  with  a  radius  equal  to 
the  distance  D  B,  describe  an  arc 
cutting  the  line  AB  at  the  points  A  and  B  ;  through  the 
point  A,  and  the  centre  D,  draw  the  diameter  AC.  Then 
through  C,  where  the  diameter  meets  the  arc,  draw  the 
straight  line  C  B,  and  it  will  be  the  perpendicular  required. 

For  the  angle  ABC,  being  inscribed  in  a  semicircle,  is 
a  right  angle  (Prop.  XVIII.  Cor.  2,  Bk.  111.)-. 


Problem  Y. 

299.  At  a  point  in  a  given  straight  line  to  make  an 
angle  equal  to  a  given  angle. 

Let  A  be  the  given 
point,  A  B  the  given  line, 
and  E  F  G  the  given 
angle. 

From  the  point  F  as 
a  centre,  with  any  radius,  describe  an  arc,  GE,  terminating 
in  the  sides  of  the  angle ;  from  the  point  A  as  a  centre, 
with  the  same  radius,  describe  the  indefinite  arc  BD. 
Draw  the  chord  G  E  ;  then  from  B  as  a  centre,  with  a 
radius  equal  to  G  E,  describe  an  arc  cutting  the  arc  B  D 
in  C.  Draw  AC,  and  the  angle  CAB  will  be  equal  to 
the  given  angle  E  F  G. 

For  the  two  arcs,  BC  and  GE,  have  equal  radii  and 
equal  chords  ;  therefore  they  are  equal  (Prop.  111.  Bk. 


BOOK   V.  121 

III.)  ;  hence  the  angles  C  A  B,  E  F  G,  measured  by  these 
arcs,  are  also  equal  (Prop.  V.  Bk.  III.). 

Problem  YI. 

300.  To  bisect  a  given  arc^  or  a  given  angle. 
First.  Let  A  D  B  be  the  given  arc  C 

which  it  is  required  to  bisect. 

Draw  the  chord  A  B  ;  from  the  cen-  / 

tre  C  draw  the  line  C  D  perpendicular  / 

to  AB  (Prob.  III.)  ;  it  will  bisect  the     a^--- 
arc  A  D  B  in  the  point  D. 

For  C  D  being  a  radius  perpendicu- 
lar to  a  chord  A  B,  must  bisect  the  arc  A  D  B  which  is 
subtended  by  that  chord  (Prop.  VI.  Bk.  III.). 

Secondly.  Let  A  C  B  be  the  angle  which  it  is  required 
to  bisect.  From  C  as  a  centre,  with  any  radius,  describe 
the  arc  A  D  B  ;  bisect  this  arc,  as  in  the  first  case,  by 
drawing  the  line  C  D  ;  and  this  line  will  also  bisect  the 
angle  ACB. 

For  the  angles  A  C  D,  BCD  are  equal,  being  measured 
by  the  equal  arcs  AD,  D  B  (Prop.  Y.  Bk.  III.). 

301.  Scholium.  By  the  sam^  construction,  we  may  bi- 
sect each  of  the  halves  AD,  D  B ;  and  thus,  by  successive 
subdivisions,  a  given  angle  or  a  given  arc  may  be  divided 
into  four  equal  parts,  into  eight,  into  sixteen,  <fec. 

Problem  YIL, 

302.  Through  a  given  point,  to  draw  a  straight  line 
parallel  to  a  given  straight  line. 

Let  A  be  the  given  point,  and     A  -^^r:::; B 

C  D  the  given  straight  line. 

From  A  draw  a  straight  line, "'•■••.., 

AE,   to   any   point,  E,  in  CD.  E 

Then  draw  A  B,  making  the  angle  E  A  B  equal  to  the 
11 


122  ELEMENTS  OF  GEOMETRY. 

angle   AEC    (Prob.  Y.)  ;   and     a^ B 

A  B  is  parallel  to  C  D.  ■"-••..... 

For  the  alternate  angles  E  AB,  ■■•••-.. 

AEC,  made  by  the  straight  line  E        ^ 

AE  meeting  the  two  straight  lines  AB,  CD,  being  equal, 
the  lines  AB  and  CD  must  be  parallel  (Prop.  XX.  Bk.  I.). 

Problem  VIII. 

303.  Tivo  angles  of  a  triangle  being  given,  to  find  the 
third  angle. 

Draw  the  indefinite  straight  line 
ABE.  At  any  point,  B,  make  the 
angle  ABC  equal  to  one  of  the 
given  angles  (Prob.  Y.),  and  the 
angle  C  B  D  equal  to  the  other  given  ABE 
angle  ;  then  the  angle  D  B  E  will  be  the  third  angle  re- 
quired. 

For  these  three  angles  are  together  equal  to  two  right 
angles  (Prop.  I.  Cor.  2,  Bk.  I.),  as  are  also  the  three  an- 
gles of  every  triangle  (Prop.  XXYIII.  Bk.  I.)  ;  and  two 
of  the  angles  at  B  having  been  made  equal  to  two  angles 
of  tlie  triangle,  the  remaini^ig  angle  D  B  E  must  be  equal 
to  the  third  angle. 

Problem  IX. 

304.  Tivo  sides  of  a  triangle  and  the  included  angle 
being  given,  to  construct  the  triangle. 

Draw  the  straight  line  A  B  equal  to 
one  of  the  two  given  sides.  At  the 
point  A  make  an  angle,  CAB,  equal  to 
the  given  angle  (Prob.  Y.)  ;  and  take 
AC  equalto  the  other  given  side.  Join  A^ 
B  C  ;  and  the  triangle  ABC  will  be  the  one  required 
(Prop.  Y.  Bk.  I.). 


BOOI^   V. 


123 


Problem  X. 

305.  One  side  and  two  angles  of  a  triangle  being  given, 
to  construct  the  triangle. 

The  two  given  angles  will  either  be  ^ 

both  adjacent  to  the  given  side,  or  one 
adjacent  and  the  other  opposite.  In 
the  latter  case,  find  the  third  angle 
(Prob.  VIII.)  ;  and  the  two  angles  ad- 
jacent to  the  given  side  will  then  be  known. 

In*the  former  case,  draw  the  straight  line  AB  eqnal  to 
the  given  side  ;  at  the  point  A,  make  an  angle,  BAG, 
equal  to  one  of  the  adjacent  angles,  and  at  B  an  angle, 
ABC,  equal  to  the  other.  Then  the  two  sides  AC,  BC 
will  meet,  and  form  with  AB  the  triangle  required  (Prop. 
VI.  Bk.  I.) 

Problem  XI. 

306.  Two  sides  of  a  triangle  and  an  angle  opposite  one 
of  them  being  given,)  to  construct  the  triangle. 

Draw  the  indefinite  straight 
line  A  B.  At  the  point  A  make 
an  angle  BAC  equal  to  the 
given  angle,  and  make  A  C 
equal  to  that  side  which  is 
adjacent  to  the  given  angle. 
Then  from  C,  as  a  centre,  with  a  radius  equal  to  the  other 
side,  describe  an  arc,  which  must  either  touch  the  line 
AB  in  D,  or  cut  it  in  the  points  E  and  F,  otherwise  a  tri- 
angle could  not  be  formed. 

When  the  arc  touches  A  B,  a  straight  line  drawn  from 
C  to  the  point  of  contact,  D,  will  be  perpendicular  to  AB 
(Prop.  XI.  Bk.  III.),  and  the  right-angled  triangle  CAD 
will  be  the  triangle  required. 

When  the  arc  cuts  A  B  in  two  points,  E  and  F,  lying 


124  ELEMENTS  OF  GEOMETRY. 

on  the  same  side  of  tlie  point  A,  draw  the  straight  lines 
C  E,  C  F  ;  and  each  of  the  two  triangles  C  A  E,  C  A  F  will 
satisfy  the  conditions  of  the  problem.  If,  however,  the 
two  points  E  and  F  should  lie  on  different  sides  of  the 
point  A,  only  one  of  the  triangles,  as  C  A  F,  will  satisfy 
all  the  conditions;  hence  that  will  be  the  triangle  required. 

307.  Scholium.  The  problem  would  be  impossible,  if 
the  side  opposite  the  given  angle  were  less  than  the  per- 
pendicular let  fall  from  the  point  C  on  the  straight  line  AB. 

Problem  XII.  « 

308.  The  three  sides  of  a  triangle  being'  given,  to  con- 
struct the  triangle. 

Draw  the  straight  line  AB  equal  to  C 

one  of  the  given  sides ;  from  the  point 
A  as  a  centre,  with  a  radius  equal  to 
either  of  the  other  two  sides,  describe  an 
arc ;.  from  the  point  B,  with  a  radius 
equal  to  the  third  side,  describe  another     ^  ^ 

arc  cutting  the  former  in  the  point  C ;  draw  the  straight 
lines  AC,  BC;  and  the  triangle  ABC  will  be  the  one 
required  (Prop.  XVIII.  Bk.  I.). 

309.  Scholium.  The  problem  would  be  impossible,  if 
one  of  the  given  sides  were  equal  to  or  greater  than  the 
sum  of  the  other  two. 

Problem  XIII. 

310.  Two  adjacent  sides  of  a  parallelogram  and  the  irir- 
eluded  angle  being  given,  to  construct  the  parallelogram. 

Draw  the  straight  line  A  B  equal 
to  one  of  the  given  sides.  At  the 
point  A  make  an  angle,  BAD,  equal 
to  the  given  angle,  and  take  AD 
equal  to  the  other  given  side.   From 


BOOK  V.  125 

the  point  D,  with  a  radius  equal  to  AB,  describe  an  arc ; 
and  from  the  point  B  as  a  centre,  with  a  radius  equal  to 
A  D,  describe  anotlier  arc  cutting  the  former  in  the  point 
C.  Draw  the  straight  lines  CD,  C  B ;  and  the  parallel- 
ogram A  B  C  D  will  be  the  one  required. 

For  the  opposite  sides  are  equal,  bj  construction  ;  hence 
the  figure  is  a  parallelogram  (Prop.  XXXII.  Bk.  I.)  ;  and 
it  is  formed  with  the  given  sides  and  the  given  angle. 

311.  Cor.  If  the  given  angle  is  a  right  angle,  the  figure 
will  be  a  rectangle  ;  and  if  the  adjacent  sides  are  albo 
equal,  the  figure  will  be  a  square. 

Problem  XIV. 

312.  A  circumference^  or  an  arc^  being  given,  to  find 
the  centre  of  the  circle. 

Take  any  three  points.  A,  B,  C, 
on  the  given  circumference,  or  arc. 
Draw  the  chords  A  B,  B  C,  and 
bisect  them  by  the  perpendiculars 
DE  and  FE  (Prob.  I.)  ;  the  point 
E,  in  which  these  perpendiculars 
meet,  is  the  centre  required. 

For  the  perpendiculars  DE,  FE 
must  both  pass  through  the  centre  (Prop.  VI.  Cor.  2, 
Bk.  III.),  and  E  being  the  only  point  through  which  they 
both  pass,  E  must  be  the  centre. 

313.  Scholium.  By  the  same  construction,  a  circumfer- 
ence may  be  made  to  pass  through  three  given  points, 
A,  B,  C,  not  in  the  same  straight  line ;  and  also  a  cir- 
cumference described  in  which  a  given  triangle,  ABC, 
shall  be  inscribed. 

Problem  XV. 

314.  T/irovgh  a  given  point  to  draw  a  tangent  to  a 

given  circle. 

11* 


126 


ELEMENTS   OP   GEOMETRY. 


First.  Let  the  given  point 
A  be  in  the  circumference. 

Find  the  centre  of  the  circle, 
C  (Prob.  XIY.)  ;  draw  the 
radius  C  A ;  flirough  the  point 
A  draw  AB  perpendicular  to 
C  A  (Prob.  IV.)  ;  and  A  B 
will  be  the  tangent  required 
(Prop.  X.  Bk.  III.). 

Secondly.  Let  the  given  point  B  be  witliont  the  circum- 
ference. 

Join  the  point  B  and  the  centre  C  by  the  straight  line 
B  C ;  bisect  B  C  in  D  ;  and  from  D  as  a  centre,  with  a 
radius  equal  to  C  D  or  D  B,  describe  a  circumference 
intersecting  the  given  circumference  in  the  points  A  and 
E.  Draw  AB  and  E  B,  and  each  will  be  a  tangent  as 
required. 

For,  drawing  C  A,  the  angle  CAB,  being  inscribed  in  a 
semicircle,  is  a  right  angle  (Prop.  XVIII.  Cor.  2,  Bk.  III.)  ; 
therefore  A  B  is  perpendicular  to  the  radius  C  A  at  its  ex- 
tremity, A,  and  consequently  is  a  tangent  (Prop.  X.  Bk. 
III.).  In  like  manner  it  may  be  shown  that  EB  is  a 
tangent. 

Phoblem  XVI. 

315.  On  a  given  straight  line  to  construct  a  segment  of 
a  circle  that  shall  contain  an  angle  equal  to  a  given  angle. 

Let  AB  be  the  given  straight 
line.  Through  the  point  B  draw 
the  straight  line  BD,  making  the 
angle  A  B  D  equal  to  the  given 
angle  ;  draw  B  E  perpendicular 
to  B  D  ;  bisect  A  B,  and  from 
F  erect  the  perpendicular  F  E. 
From  the  point  E,  where  these 
perpendiculars  meet,  as  a  centre. 


with  the  distance  EB 


BOOK  V.  127 

or  E  A,  describe  a  circumference,  and  A  C  B  will  be  the 
segment  rcqnircd. 

For,  since  B  D  is  a  perpendicular  at  the  extremity  of 
the  radius  E  B,  it  is  a  tangent  (Prop.  X.  Bk.  III.)  ; 
and  the  angle  A  B  D  is  measured  by  half  the  arc  A  G  B 
(Prop.  XX.  Bk.  III.).  Also,  the  angle  ACB,  being  an 
inscribed  angle,  is  measured  by  half  the  arc  A  G-  B  ;  there- 
fore the  angle  ACB  is  equal  to  the  angle  A BD.  But, 
by  construction,  the  angle  ABD  is  equal  to  the  givea 
angle  ;  hence  the  segment  ACB  contains  an  angle  equal 
to  the  given  angle. 

316.  Scholium,  If  the  given  angle  were  acute,  the 
centre  must  lie  within  the  segment  (Prop.  XYIII.  Cor.  3, 
Bk.  III.)  ;  and  if  it  were  right,  the  centre  must  be  in  the 
middle  of  the  line  A  B,  and  the  required  segment  would 
be  a  semicircle. 

Problem  XVII. 

317.  To  inscribe  a  circle  in  any  given  triangle. 
Bisect  any  two  of  the  angles, 

as  A  and  B,  by  the  straight 
lines  A  E  and  B  E,  meeting  in 
the  point  E  (Prob.  YI.) .  From 
the  point  E  let  fall  the  perpen- 
diculars ED,  EF,  EG  (Prob. 
II.)  on  the  three  sides  of  the 
triangle  ;  these  perpendiculars 
will  all  be  equal. 

For,  by  construction,  we  have  the  angle  DAE  equal  to 
the  angle  EAG,  and  the  right  angle  ADE  equal  to  the 
right  angle  AGE;  hence  the  third  angle  A  E  D  is  equal 
to  the  third  angle  A  E  G.  Moreover,  A  E  is  common  to 
the  two  triangles  A  ED,  AEG;  hence  the  triangles  them- 
selves are  equal,  and  E  D  is  equal  to  E  G.  In  the  same 
manner  it  may  be  shown  that  the  two  triangles  BEF, 


128  ELEMENTS  OP  GEOMETRY. 

BEG  are  equal ;  therefore  E  F  is  equal  to  E  G ;  hence 
tlie  three  perpendiculars  ED,  E  F,  E  G  are  all  equal,  and 
if,  from  the  point  E  as  a  centre,  with  the  radius  ED,  a 
circle  be  described,  it  must  pass  througli  the  points  F  and  G. 

318.  Scholium.  The  three  lines  whicli  bisect  the  angles 
of  a  triangle  all  meet  in  the  centre  of  the  inscribed  circle. 

Problem  XYIII. 

319.  To  inscribe  a  circle  in  a  given  square. 
Draw  the   diagonals  AC,  D  B,  and 

from  the  point  E,  where  the  diagonals 
mutually  bisect  each  other  (Prop. 
XXXIV.  Bk.  I.),  draw  the  straight  line 
EF  perpendicular  to  a  side  of  the  square. 
From  E  as  a  centre,  with  a  radius  equal 
to  EF,  describe  a  circle,  and  it  will 
touch  each  side  of  the  square. 

For  the  square  is  divided  by  the  diagonals  into  four 
equal  isosceles  triangles  ;  hence,  the  perpendicular,  from 
the  vertex  E  to  the  base,  is  the  same  in  each  triangle  ; 
therefore  the  circumference  described  from  the  centre  E, 
with  the  radius  E  F,  passes  through  the  extremities  of 
each  perpendicular  ;  consequently,  the  sides  of  the  square 
are  tangents  to  the  circle  (Prop.  X.  Bk.  III.). 

Problem  XIX. 

320.  To  find  the  side  of  a  square  tvhich  shall  be  equiv- 
alent to  the  sum  of  two  given  squares. 

Draw  the  two  straight  lines  A  B,  B  C 
perpendicular  to  each  other,  taking  A  B 
equal  to  a  side  of  one  of  the  given  squares, 
and  B  C  equal  to  a  side  of  tlic  other. 
Join  AC;  this  will  be  the  side  of  the 
square  required. 


BOOK   V, 


129 


For,  the  triangle  ABC  being  riglit-angled,  the  square 
that  can  be  described  upon  the  hypothenuse  A  C  is  equiv- 
alent to  tlie  sum  of  the  squares  that  can  be  described  upon 
tlie  sides  A  B  and  B  C  (Prop.  XI.  Bk.  lY.). 

321.  Scholium,  A  square  may  thus  be  found  equivalent 
to  the  sum  of  any  number  of  squares  ;  for  the  construc- 
tion which  reduces  two  of  them  to  one,  will  reduce  three 
of  them  to  two,  and  these  two  to  one. 


Problem  XX. 

322.  To  find  the  side  of  a  square  vjhich  shall  be  equiv- 
alent to. the  difference  of  tuw  given  squares. 

Draw  the  two  straight  lines  A  B,  A  C  C 
perpendicular  to  each  other,  making 
A  C  equal  to  the  side  of  the  less  square. 
Then  from  C  as  a  centre,  with  a  radius 
equal  to  tlie  side  of  the  otlier  square, 
describe  an  arc  intersecting  A  B  in  the 
point  B,  and  A  B  will  be  the  side  of  the  required  square. 

For,  join  B  C,  and  the  square  that  can  be  described 
upon  AB  is  equivalent  to  the  difference  of  the  squares 
that  can  be  described  on  B  C  and  A  C  (Prop.  XI.  Cor.  1, 
Bk.  IV.). 

Problem  XXI. 

323.  To  construct  a  rectangle  that  shall  he  equivalent 
to  a  given  triangle. 

Let  ABC  be  the  given  triangle. 

Draw  the  indefinite  straight  line 
C  D  parallel  to -the  base  AB  ;  bi- 
sect A  B  by  the  perpendicular 
EF,  and  make  E  G  equal  to  F  B. 
Then,  by  drawing  G  B,  tlie  rec- 
tangle E  F  B  G  is  equal  to  the  tri- 
angle ABC. 


ISO  ELEMENTS  OF  GEOMETRY. 

For  the  rectangle  EFBG  has  the  same  altitude,  EF,  as 
the  triangle  ABC,  and  half  its  base  (Prop.  II.  Cor.  1, 
Bk.  lY.). 

Problem  XXII. 

324.  To  construct  a  triangle  that  shall  he  equivalent  to 
a  given  -polygon. 

Let  A  B  C  D  E  be  the  given  poly- 
gon. 

Draw  the  diagonal  CE,  cutting  off 
the  triangle  C  D  E  ;  through  the 
point  D  draw  D  F  parallel  to  C  E, 
and  meeting  A  E  produced  in  F.  ^  -^  ^  ^ 
Draw  C  F  ;  and  tlie  polygon  A  B  C  D  E  will  be  equivalent 
to  the  polygon  A  B  C  F,  whicli  has  one  side  less  than  the 
given  polygon. 

For  the  triangles  C  D  E,  C  F  E  have  the  base  C  E  com- 
mon ;  they  have  also  the  same  altitude,  »ince  their  ver- 
tices, D,  F,  are  situated  in  a  line,  DF,  parallel  to  the  base  ; 
these  triangles  are  therefore  equivalent  (Prop.  II.  Cor.  2, 
Bk.  lY.).  Add  to  each  of  them  the  figure  AB  C  E,  and 
the  polygon  A  B  C  D  E  will  be  equivalent  to  the  polygon 
A  B  C  F. 

In  like  manner,  the  triangle  C  G  A  may  be  substituted 
for  tlie  equivalent  triangle  ABC,  and  tluis  the  pentagon 
A  B  C  D  E  will  be  changed  into  an  equivalent  triangle 
GCF. 

The  same  process  may  be  applied  to  every  other  polygon  ; 
for,  by  successively  diminishing  the  number  of  its  sides, 
one  at  each  step  of  the  process,  the  equivalent  triangle 
will  at  last  be  found.  * 

Problem  XXIII. 

325.  To  divide  a  given  straight  line  into  any  number 
of  equal  parts* 


BOOK   V. 


131 


Let  A  B  be  the  given  straight 
line  proposed  to  be  divided 
into  any  number  of  equal  parts ; 
for  example,  six. 

Through   the   extremity   A 
draw    the    indefinite    straight 

line  A  E,  making  any  angle  with  A  B.  Take  A  C  of  any 
convenient  length,  and  apply  it  six  times  upon  A  E.  Join 
the  last  point  of  division,  E,  and  the  extremity  B  by  tlie 
straight  line  E  B  ;  and  through  the  point  C  draw  C  D  par- 
allel to  E  B  ;  then  A  D  will  be  the  sixth  part  of  the  line 
A  B,  and,  being  applied  six  times  to  A  B,  divides  it  into 
six  equal  parts. 

For,  since  C  D  is  parallel  to  E  B,  in  the  triangle  ABE, 
we  have  the  proportion  (Prop.  XVII.  Bk.  IV.)) 

AD  :  AB  :  :  AC:  AE. 
But  A  C  is  the  sixth  part  of  A  E  ;  hence  A  D  is  the  sixth 
part  of  A  B. 

Problem  XXIV. 

32G.  To  divide  a  given  straight  line  into  parts  that  shall 
he  proportional  to  other  given  lines. 

Let  A  B  be  the  given  straight 
line  proposed  to  be  divided  into 
parts  proportional  to  the  given 
lines  AC,  CD,  D  E. 

Through  the  point  A  draw  the 
indefinite  straight  line  AE,  mak- 
ing any  angle  with  A  B.  On  A  E  lay  off  A  C,  C  D,  and 
D  E.  Join  the  points  E  and  B  by  the  straight  line  E  B, 
and  through  the,  points  C  and  D  draw  C  G  and  D  H  par- 
allel to  E  B ;  and  the  line  A  B  will  be  divided  into  parts 
proportional  to  the  given  lines. 

For,  since  C  G  and  D  H  are  each  parallel  to  E  B,  wo 
have  the  proportion  (Prop.  XVII.  Cor.  2,  Bk.  IV.), 
A  C  :  A  G  :  :  C  D  :  G  H  :  :  D  E  :  H  B. 


182  ELE^IENTS   OF   GEOMETRY 


Problem  XXV. 


827.    To  find  a  fourth  proportional  to   three  given 
straight  lines. 

Draw  the  two  indefinite  straight 
lines  AB,  AE,  forming  any  angle 
with  each  other. 

On  AB  make  AD  equal  to  the 
first  of  the  proposed  lines,  and  A  B 
equal  to  the  second  ;  and  on  A  E     A  D  B 

make  AE  equal  to  the  third.  Join  BE ;  and  through  the 
point  D  draw  D  C  parallel  to  B  E,  and  A  C  will  be  the 
fourth  proportional  required. 

For,  since  D  C  is  parallel  to  B  E,  we  have  the  propor- 
tion (Prop.  XVII.  Cor.  1,  Bk.  IV.), 

AB  :  AD  :  :  AE  :  AC. 

828.  Cor.  A  third  proportional  to  two  given  lines,  A 
and  B,  may  be  found  in  the  same  manner,  for  it  will  be 
the  same  as  a  fourth  proportional  to  the  three  lines, 
A,  B,  and  B. 

Problem  XXVI. 

329.  To  find  a  mean  proportional  between  tioo  given 
straight  lines. 

Draw  the  indefinite  straight  line 
AB.     On   AB   take   AC  equal  to  ^ 

the  first  of  the  given  lines,  and  C  B 
equal  to  the  second.  On  A  B,  as  a 
diameter,  describe  a  semicircle,  and 
at  the  point  C  draw  the  perpendic- 
ular C D,  meeting  the  semi-circumference  in  D  ;  CD  will 
be  the  mean  proportional  rcfquired. 

For  the  perpendicular  C  D,  drawn  from  a  point  in  the 
circumference  to  a  point  in  the  diameter,  is  a  mean  pro- 


B 


BOOK   V.  133 

portional  between  the  two  segments  of  the  diameter  A  C, 
C  B  (Prop.  XXXII.  Cor.,  Bk.  lY.)  ;  and  these  segments 
are  equal  to  the  given  lines. 

Problem  XXYII. 

330.  To  divide  a  given  straight  line  into  two  such  parts^ 
that  the  greater  part  shall  be  a  mean  proportional  betioeen 
the  whole  line  and  the  other  part. 

Let  A  B  be  the  given  straight  ^ ^.^ 

line.  /  \j« 

At  the  extremity,  B,  of  the  line  / 

AB,  erect  the  perpendicular  BC,  1.         ^^ 

equal  to  the  half  of  A  B.     From  ^^\ 

the  point  C  as  a  centre,  with  the 


radius    C  B,    describe    a   circle.     -^  E        B 

Draw  AC  cutting  the  circumference  in  D  ;  and  take  AE 
equal  to  A  D.  The  line  A  B  will  be  divided  at  the  point 
E  in  the  manner  required  ;  that  is, 

AB:  AE  :  :  AE.EB. 

For  A  B,  being  perpendicular  to  the  radius  at  its  ex- 
tremity, is  a  tangent  (Prop.  X.  Bk.  III.)  ;  and  if  A  C  be 
produced  till .  it  again  meets  the  circumference,  in  F,  we 
shall  have  (Prop.  XXXV.  Bk.  IV.), 

AF:  AB:  :  AB:  AD; 
hence,  by  division  (Prop.  VIII.  Bk.  II.), 

AF  —  AB:AB::AB  —  AD:AD. 

But,  since  the  radius  is  the  half  of  A  B,  the  diameter 
D  F  is  equal  to  A  B,  and  consequently  A  F  —  A  B  is  equal 
to  AD,  which  is  equal  to  A E  ;  also,  since  A E  is  equal  to 
A  D,  we  have  A  B  —  AD  equal  to  E  B  ;  hence, 

AE  :  AB  :  :  EB  :  AD,  or  AE; 
and,  by  inversion  (Prop.  V.  Bk.  II.), 

AB:  AE:  :  AE:EB. 

12 


134 


ELEMENTS   OF   GEOMETRY. 


331.  Scholium.  This  sort  of  division  of  the  line  A  B  is 
called  division  in  extreme  and  mean  ratio. 

Problem  XXVIII. 

332.  Tlirough  a  given  point  in  a  given  angle,  to  draio 
a  straight  line,  ivhich  shall  have  the  parts  included  be- 
tiveen  that  point  and  the  sides  of  the  angle  equal  to  each 
other. 

Let  E  be  the  given  point,  and  ABC 
the  given  angle. 

Through  the  point  E  draw  E  F  paral- 
lel to  BC,  make  AF  equal  to  BF. 
Through  the  points  A  and  E  draw  the 
straight  line  A  E  C,  and  it  will  be  the 
line  required. 

For,  E  F  being  parallel  to  B  C,  we  have  (Prop.  XYII. 
Bk.  lY.), 

AF:FB:  :  AE:EC; 

but  A  F  is  equal  to  F  B ;  therefore  A  E  is  equal  to  E  0. 

Problem  XXIX. 

333.  On  a  given  straight  line  to  construct  a  rectangle 
that  shall  be  equivalent  to  a  given  rectangle. 

Let  AB  be  the  given  straight  p  E 

line,  and  CDEF  the   given     H  G 

rectangle. 

Find  a  fourth  proportional 
to  the  three  straiglit  lines 
AB,  CD,DE  (Prob.XXY.); 
and  let  B  G  be  that  fourth  proportional.  The  rectangle 
constructed  on  A  B  and  B  G  will  be  equivalent  to  the  rec- 
tangle CDEF. 

For,  since  AB  :  CD  :  :  D  E:  B  G,  it  follows  (Prop. 
I.  Bk.  II.)  that 

AB  X  BG=CD  X  DE; 


BOOK   V. 


135 


hence,  the  rectangle  A  B  G  H,  which  is  constructed  on  the 
line  A  B,  is  equivalent  to  the  rectangle  C  D  E  F. 

Problem  XXX. 

334.    To  construct  a  square  that  shall  be  equivalent  to  a 
given  parallelogram,  or  to  a  given  triangle. 

First.  Let   A.B  C  D   be   the   given  D  C 

parallelogram,  A  B  its  base,  and  D  E  / 

its  altitude.  / 

Find  a  mean  proportional  between       LA 

A  B  and  D  E    (Frob.  XXYI.)  ;  and      ^   ^ 
the  square  constructed  on  that  proportional  will  be  equiv- 
alent to  the  parallelogram  A  B  C  D. 

For,  denoting  the  mean  proportional  by  xy,  we  have, 
by  construction, 

A  B  :  a;  y  :  :  a;  y  :  D  E  ; 
therefore, 

5c]/^  =  A  B  X  D  E ; 

but  A  B  X  B  E  is  the  measure  of  the  parallelogram,  and 
X  y  that  of  the  square  ;  hence  they  are  equivalent. 

Secondly.  Let  A  B  C  be  the  given  tri- 
angle, BC  its  base,  and  AD  its  altitude. 

Find   a   mean   proportional    between  / 

BC  and  the  half  of  AD,  and  let  xy  / 

denote   that  proportional  ;    the    square       / 
constructed  on  xy   will  be  equivalent     B  DC 

to  the.triangle  ABC. 

For  since,  by  construction, 

B  C  :  xy  :  :  xy  :  j  A D, 
it  follows  that 

5r^^=BC  X  ^  AD; 

hence  the  square  constructed  on  xy  is  equivalent  to  the 
triangle  ABC. 


136  ELEMENTS  OP  GEOMETRY. 


Problem  XXXI. 

335.  To  construct  a  rectangle  equivalent  to  a  given 
square,  and  having  the  sum  of  its  adjacent  sides  equal  to 
a  given  line. 

Let  the  straight  line  A  B  be  equal 
to  the  sum  of  the  adjacent  sides  of 


/ 


the  required  rectangle. 

Upon  A  B  as  a  diameter  describe 
a  semicircle  ;  at  the  point  A,  draw    ^  E    B 

A  D  perpendicular  to  A  B,  making  A  D  equal  to  the  side 
of  the  given  square  ;  then  draw  the  line  D  C  parallel  to 
the  diameter  A  B.  From  the  point  C,  where  the  parallel 
meets  the  circumference,  draw  C  E  perpendicular  to  the 
diameter  ;  A  E  and  E  B  will  be  the  sides  of  the  rectangle 
required. 

For  their  sum  is  equal  to  A  B  ;  and  their  rectangle 
A  E  X  E  B  is  equivalent  to  the  square  of  C  E,  or  to  the 
square  of  A  D  (Prop.  XXXII.  Cor.,  Bk.  lY.)  ;  hence, 
this  rectangle  is  equivalent  to  the  given  square. 

336.  Scholium.  The  problem  is  impossible,  when  the 
distance  A  D  is  greater  than  the  half  the  given  line  A  B, 
for  then  the  line  D  C  will  not  meet  the  circumference. 

Problem  XXXII. 

337.  To  construct  a  rectangle  that  shall  be  equivalent 
to  a  given  square,  and  the  difference  of  whose  adjacent 
sides  shall  be  equal  to  a  given  line. 

Let  the  straight  line  A  B  be  equal  to  the  difference  of 
the  adjacent  sides  of  the  required  rectangle. 

Upon  A  B  as  a  diameter,  describe  a  circle.  At  the  ex- 
tremity of  the  diameter,  draw  the  tangent  AD,  making  it 
equal  to  the  side  of  the  given  square. 


BOOK   V. 


137 


Through  the  point  D  and  the  centre 
C  draw  the  secant  D  C  F,  intersecting 
the  circumference  in  E ;  then  D  E 
and  D  F  will  be  the  adjacent  sides  of 
the  rectangle  required. 

For  the  difference  of  these  lines  is 
equal  to  the  diameter  E  F  or  A  B  ; 
and  the  rectangle  D  E  X  D  F  is  equal 
to  AD^  (Prop.  XXXY.  Cor.,  Bk.  IV.)  ;  hence  it  is  equiv- 
alent to  the  given  square. 


Problem  XXXIII. 

338.   To  construct  a  square  that  shall  be  to  a  given 
square  as  one  given  line  is  to  another  given  line. 

Draw  the  indefinite  line  AB, 
on  which  take  A  C  equal  to  one 
of  the  given  lines,  and  C  B  equal 
to  the  other.  Upon  A  B  as  a  di- 
ameter, describe  a  semicircle,  and 
at  the  point  C  draw  the  perpendicular  C  D,  meeting  the 
circumference  in  D.  Through  the  points  A  and  B  draw 
the  straight  lines  D  E,  D  F,  making  the  former  equal  to 
the  side  of  the  given  square  ;  and  through  the  point  E 
draw  E  F  parallel  to  A  B ;  D  F  will  be  the  side  of  the 
square  required. 

For,  since  E  F  is  parallel  to  A  B, 

DE  :  DF:  :  D  A:  DB; 
consequently  (Prop.  XY.  Bk.  II.), 

5^^  :  DF^  :  :  D^  :  D'bI 

But  in  the  right-angled  triangle  A  D  B  the  square  of  A  D 
is  to  the  square  of  D  B  as  the  segment  A  C  is  to  the  seg- 
ment C  B  (Prop.  XI.  Cor.  3,  Bk.  TV.)  ;  hence, 

DE^  DF':  :  AC:  CB. 

12* 


ld»  ELEMENTS   OP   GEOMETRY. 

But,  by  construction,  D  E  is  equal  to  the  side  of  the  given 
square  ;  also,  A  C  is  equal  to  one  of  the  given  lines,  and 
C  B  to  the  other  ;  hence,  the  given  square  is  to  that  con- 
structed on  D  F  as  the  one  given  line  is  to  the  other. 

Problem  XXXIY. 

339.  Upon  a  given  base  to  construct  an  isosceles  tri- 
angle^ having  each  of  the  angles  at  the  base  double  the 
vertical  angle. 

Let  A  B  be  the  given  base. 

Produce  A  B  to  some  point  C  till  the  -^ 

rectangle  A  C  X  B  C  shall  be  equiva-  A 

lent    to    the   square   of   A  B    (Prob.  /    \ 

XXXII.)  ;  then,  with  the  base  A  B  and  y v     \ 

sides  each  equal  to  A  C,  construct  the  /     ^sV 

isosceles  triangle  DAB,  and  the  angle  A  B   C 

A  will  double  the  angle  D. 

For,  make  DE  equal  to  AB,  or  make  AE  equal  to  BC, 
and  join  E  B.     Then,  by  construction, 

AD  :  AB:  :  AB:  AE; 

for  AE  is  equal  to  BC  ;  consequently  the  triangles  DAB, 
B  A  E  have  a  common  angle.  A,  contained  by  proportional 
sides  ;  hence  they  are  similar  (Prop.  XXIY.  Bk.  IV.)  ; 
therefore  these  triangles  are  both  isosceles,  for  D  A  B  is 
isosceles  by  construction,  so  that  A  B  is  equal  to  E  B ;  but 
A  B  is  equal  to  D  E  ;  consequently  D  E  is  equal  to  E  B, 
and  therefore  the  angle  D  is  equal  to  the  angle  E  B  D  ; 
hence  the  exterior  angle  A  E  B  is  equal  to  double  the  an- 
gle D,  but  the  angle  A  is  equal  to  the  angle  AEB ;  there- 
fore the  angle  A  is  double  the  angle  D. 

Problem  XXXY. 

340.  Upon  a  given  straight  line  to  construct  a  polygon 
similar  to  a  given  polygon. 


BOOK  V.  139 

Let  ABC  DE  C 

be  the  given  poly-            ^^^'^7^\  H 

ffon,  and  F  G  the     ^^f"^  /.        ^\  .^^<^ 

given  straight  line.        \     /'         -/-^      \     ..-■■'  \ 

Draw  the  diag-     ^  liij^  /^  fV / 

onals    AC,   AD.  ^^^^^.X  \^X 

At  the  point  F  in  ^  ^ 

the  straight  line  F  G,  make  the  angle  G  F  H  equal  to  the 
angle  B  A  C  ;  and  at  the  point  G  make  the  angle  F  G  H 
equal  to  the  angle  ABC.  The  lines  FH,  GPI  will  cut 
each  other  in  H,  and  FGH  will  be  a  triangle  similar  to 
ABC.  In  the  same  manner,  upon  F  H,  homologous  to 
A  C,  construct  the  triangle  F  I  H  similar  to  A  D  C  ;  and 
upon  FI,  homologous  to  AD,  construct  the  triangle  FIK 
similar  to  A  D  E.  The  polygon  F  G  H I K  will  be  similar 
to  A  B  C  D  E,  as  required. 

For  these  two  polygons  are  composed  of  the  same  num- 
ber of  triangles,  similar  each  to  each,  and  similarly  situ- 
ated (Prop.  XXX.  Cor.,  Bk.  lY.). 

Problem  XXXYI. 

341.  Two  similar  polygons  being  given,  to  construct  a 
similar  polygon,  which  shall  be  equivalent  to  their  sum  or 
their  differetice. 

Let  A  and  B  be 
two  homologous  sides 
of  the  given  poly- 
gons. 

Find  a  square  equal 
to  the  sum  or  to 
the  difference  of  tlie 
squares  described  up- 
on A  and  B  ;  let  x  be  the  side  of  that  square ;  then  will  x 
in  the  polygon  required  be  the  side  which  is  homologous 
to  the  sides  A  and  B  in  tlie  given  polygons.  The  polygon 
itself  may  then  be  constructed  on  x,  by  the  last  problem. 


140  ELEMENTS   OF   GEOMETRY. 

For  similar  figures  are  to  each  other  as  the  squares  of 
their  homologous  sides  ;  but  the  square  of  the  side  x  is 
equal  to  the  sum  or  the  difference  of  the  squares  described 
upon  the  homologous  sides  A  and  B ;  therefore  the  figure 
described  upon  the  side  x  is  equivalent  to  the  sum  or  to 
the  difference  of  the  similar  figures  described  upon  the 
sides  A  and  B. 

Problem  XXXYII. 

342.  To  construct  a  polygon  similar  to  a  given  polygon^ 
and  which  shall  have  to  it  a  given  ratio. 

Let  A  be  a  side  of  the  given  polygon. 

Find  the  side"B  of  a  square,  which  is 
to  the  square  on  A  in  the  given  ratio 
of  the  polygons  (Prob.  XXXIIL). 

Upon  B  construct  a  polygon  similar 
to  the  given  polygon  (Prob.  XXXV.), 
and  B  will  be  the  polygon  required. 

For  the  similar  polygons  constructed  upon  A  and  B 
have  the  same  ratio  to  each  other  as  the  squares  con- 
structed upon  A  and  B  (Prop.  XXXI.  Bk.  IV.). 

Problem  XXXVIII. 

343.  To  construct  a  polygon  similar  to  a  given  polygon, 
P,  and  which  shall  be  equivalent  to  another  polygon,  Q. 

Find  M,  the  side  of  a  square, 
equivalent  to  the  polygon  P, 
and  N,  the  side  of  a  square 
equivalent  to  the  polygon  Q. 
Let  a;  be  a  fourth  propor- 
tional to  the  three  given  lines 
M,  N,  A  B  ;  vipon  the  side  x,  homologous  to  A  B,  describe 
a  polygon  similar  to  the  polygon  P  (Prob.  XXXV.)  ;  it 
will  also  be  equivalent  to  the  polygon  Q. 


BOOK   Y.  Ml 

For,  representing  the  polygon  described  upon  the  side 
xhj  y^  we  have 

P  :  y  :  :  AB'  :  x" ; 

but,  by  construction, 

A B  :  rr  :  :  M  :  N,     or     AB^  :  x" :  :  M^  :  N^; 

hence. 

But,  by  construction  also,  M^  is  equivalent  to  P,  and 
N^  is  equivalent  to  Q ;  therefore, 

P:7/::P:Q; 

consequently  1/  is  equal  to  Q  ;  hence  the  polygon  1/  is 
similar  to  the  polygon  P,  and  equivalent  to  the  poly- 
gon Q. 


BOOK   VI 


REGULAR  POLYGONS,  AND    THE    AREA   OF   THE 
CIRCLE. 


DEFINITIONS. 

344.  A  Regular  Polygon  is  one  which  is  both  equi- 
lateral and  equiangular. 

345.  Regular  polygons  may  have  any  number  of  sides  : 
the  equilateral  triangle  is  one  of  three  sides  ;  the  square 
is  one  of  four. 

Proposition  I. — Theorem. 

346.  Regular  polygons  of  the  same  number  of  sides  are 
similar  figures. 

LetABCDEF,  E  D 

GHIKLM,  be 
two  regular  poly- 
gons of  the  same 
number  of  sides ; 
then  these  poly- 
gons  are   similar.  A  B  G        H 

For,  since  the  two  polygons  have  the  same  number  of 
sides,  they  have  the  same  number  of  angles  ;  and  the  sum 
of  all  the'  angles  is  the  same  in  the  one  as  in  the  other 
(Prop.  XXIX.  Bk.  I.).  Also,  since  the  polygons  are 
equiangular,  each  of  the  angles  A,  B,  C,  <fec.  is  equal  to 
each  of  the  angles  G,  H,  I,  <fec. ;  hence  the  two  polygons 
are  mutually  equiangular. 


BOOK  VI.  .  143 

Again  ;  the  polygons  being  regular,  the  sides  A  B,  B  C, 
CD,  &G.  are  equal  to  each  other;  so  likewise  are  the  sides 
G  H,  H  I,  I K,  &c.     Hence, 

AB:GH::BC:HI::CD:IK,  &c. 

Therefore  the  two  polygons  have  their  angles  equal,  and 
their  homologous  sides  proportional ;  hence  they  are  simi- 
lar (Art.  210). 

347.  Cor.  The  perimeters  of  two  regular  polygons  of 
the  same  number  of  sides,  are  to  each  other  as  their  ho- 
mologous sides,  and  their  areas  are  to  each  other  as  the 
squares  of  those  sides  (Prop.  XXXI.  Bk.  IV.). 

348.  Scholium.  The  angle  of  a  regular  polygon  js  de- 
termined by  the  number  of  its  sides  (Prop.  XXIX.  Bk.  I.). 

Proposition  II.  —  Theorem. 

349.  A  circle  map  be  circumscribed  about,  and  another 
inscribed  in,  any  regular  polygon. 

Let  ABCDEFGH  be  any  reg- 
ular polygon  ;  then  a  circle  may  be 
circumscribed  about,  and  another 
inscribed  in  it. 

Describe  a  circle  whose  circum- 
ference shall  pass  through  the  three 
points  A,  B,  C,  the  centre  being  0  ; 
let  fall  the  perpendicular  0  P  from ' 
0  to  the  middle  pohit  of  the  side  B  C  ;  and  draw  the 
straight  lines  0  A,  OB,  0  C,  OD. 

Now,  if  the  quadrilateral  0  P  C  D  be  placed  upon  the 
quadrilateral  OP  B  A,  they  will  coincide  ;  for  the  side  OP 
is  common,  and  the  angle  0  P  C  is  equal  to  the  angle 
OPB,  each  being  a  right  angle;  consequently  the  side 
P  C  will  fall  upon  its  equal,  P  B,  and  the  point  C  on  B. 
Moreover,  from  tlie  nature  of  the  polygon,  the  angle  PCD 
is  equal  to  the  angle  PB  A  ;  therefore  C  D  will  take  the 


144  .ELEMENTS  OF  GEOMETRY. 

direction  B  A,  and  C  D  being  equal 
to  B  A,  tlie  point  D  will  fall  upon 
A,  and  the  two  quadrilaterals  will 
coincide  throughout.  Therefore 
OD  is  equal  to  AO,  and  the  circum- 
ference which  passes  through  the 
three  points  A,  B,  C,  will  also  pass 
through  the  point  D.  By  the  same 
mode  of  reasoning,  it  may  be  shown  that  the  circle  which 
passes  through  the  three  vertices  B,  C,  D,  will  also  pass 
through  the  vertex  E,  and  so  on.  Hence,  the  circumfer- 
ence which  passes  through  the  three  points  A,  B,  C,  passes 
through  the  vertices  of  all  the  angles  of  the  polygon,  and 
is  circumscribed  about  the  polygon  (Art.  166). 

Again,  with  respect  to  this  circumference,  all  the  sides, 
AB,  B  C,  CD,  <fec.,  of  the  polygon  are  equal  chords ;  con- 
sequently they  are  equally  distant  from  the  centre  (Prop. 
YIII.  Bk.  III.).  Hence,  if  from  the  point  0,  as  a  centre, 
and  with  the  radius  0  P,  a  circle  be  described,  the  circum- 
ference will  touch  the  side  B  C,  and  all  the  other  sides  of 
the  polygon,  each  at  its  middle  point,  and  the  circle  will 
be  inscribed  in  the  polygon  (Art.  168). 

350.  Scholium  1.  The  point  0,  the  common  centre  of 
the  circumscribed  and  inscribed  circles,  may  also  be  re- 
garded as  the  centre  of  the  polygon.  The  angle  formed 
at  the  centre  by  two  radii  drawn  to  the  extremities  of  the 
same  side  is  called  the  angle  at  the  centre  ;  and  the  per- 
pendicular from  the  centre  to  a  side  is  called  the  apothegm 
of  the  polygon. 

Since  all  the  chords  AB,  B  C,  CD,  <fec.  are  equal,  all 
the  angles  at  the  centre  must  likewise  be  equal ;  therefore 
tlie  value  of  each  may  be  found  by  dividing  four  right  an- 
gles by  the  number  of  sides  of  the  polygon. 

351.  Scholium  2.  To  inscribe  a  regular  polygon  of  any 
number  of  sides  in  a  given  circle,  it  is  only  necessary  to 


BOOK   VI.  145 

divide    the    circumference    into    as 

many  equal   parts   as   the   polygon 

has  sides  ;  for  the  arcs  being  equal, 

the  chords  AB,  BC,  CD,  &c.  are 

also    equal    (Prop.  III.  Bk.  III.)  ; 

hence  likewise  the  triangles  A  0  B, 

B  0  C,  COD,  &c.  must  be  equal,  since  their  sides  are 

equal  each  to  each  (Prop.  XVIII.  Bk.  I.)  ;  therefore  all 

the  angles  ABC,  BCD,  CDE,  &c.  are  equal ;  hence 

the  figure  ABCDEF  is  a  regular  polygon. 

Proposition  III.  —  Theorem. 

352.  If  from  a  common  centre  a  circle  can  he  circum- 
scribed about,  and  another  circle  inscribed  within,  a  poly' 
gon,  that  polygon  is  regular . 

Suppose  that  from  the  point  0, 
as  a  centre,  circles  can  be  circum- 
scribed about,  and  inscribed  in,  the 
polygon  ABCDEF;  then  that 
polygon  is  regular. 

For,  supposing  it  to  be  described, 
the   inner   one  will   touch   all   the  A  B 

sides  of  the  polygon  ;  therefore  these  sides  are  equally  dis- 
tant from  its  centre ;  and  consequently,  being  chords  of 
tlie  outer  circle,  they  are  equal ;  therefore  tliey  include 
equal  angles  (Prop.  XVIII.  Cor.  1,  Bk.  III.).  Hence 
the  polygon  is  at  once  equilateral  and  equiangular ;  con- 
sequently it  is  regular  (Art.  344). 

Proposition  IV. — Problem. 

353.  To  inscribe  a  square  in  a  given  circle. 

Draw  two  diameters,  AC,  BD,  intersecting  each  other 
at  riglit  angles  ;  join  their  extremities,  A,  B,  C,  D,  and 
the  figure  A  B  C  D  will  be  a  square. 

13 


146 


ELEMENTS   OF    GEOMETRY.' 


For,  the  angles  A  OB,  BOC,  &c.  be- 
ing equal,  the  chords  AB,  BC,  <fec.  are 
also  equal  (Prop.  Ill,  Bk.  III.)  ;  and 
the  angles  ABC,  BCD,  <fec.,  being  in- 
scribed in  semicircles,  are  right  angles 
(Prop.  XVIII.  Cor.  2,  Bk.  III.).  Hence 
A  B  C  D  is  a  square,  and  it  is  inscribed  in  the  circle 
ABCD. 

354.    Cor.    Since  the  triangle  A  0  B  is  right-angled  and 
isosceles,  we  have  (Prop.  XI.  Cor.  5,  Bk.  IV.), 

AB:  AO:  :  V2  :  1; 

hence,  the  side  of  the  inscribed  square  is  to  the  radius  as 
the  square  root  of  2  is  to  unity. 


Proposition  V.  —  Theorem. 

355.    The  side  of  a  regular  hexagon  inscribed  in  a 
circle  is  equal  to  the  radius  of  the  circle. 

Let  ABCDEF  be  a  regular 
hexagon  inscribed  in  a  circle,  the 
centre  of  which  is  0 ;  then  any 
side,  as  B  C,  will  be  equal  to  the 
radius  OA. 

Join  BO;  and  the  angle  at  the 
centre,  A  OB,  is  one  sixth  of  four 
right  angles  (Prop.  II.  Sch.  1),  or  one  third  of  two  right 
angles  ;  therefore  the  two  other  angles,  0  A  B,  0  B  A,  of 
the  same  triangle,  are  together  equal  to  two  tliirds  of  two 
right  angles  (Prop.  XXVIII.  Bk.  I.).  But  A  0  and  B  O 
being  equal,  the  angles  OAB,  OB  A  are  also  equal  (Prop. 
VII.  Bk.  I.)  ;  consequently,  each  is  one  third  of  two  right 
angles.  Hence  the  triangle  A  0  B  is  equiangular  ;  there- 
fore A  B,  the  side  of  the  regular  hexagon,  is  equal  to  A  0, 
the  radius  of  the  circle  (Prop.  VIII.  Cor.  Bk.  I.). 


BOOK    VI. 


147 


356.  Cor.  1.  To  inscribe  a  regular 
hexagon  iii  a  given  circle,  apply  the  ra- 
dius, A  0,  of  the  circle  six  times,  as  a 
chord  to  the  circumference.  Hence, 
beginning  at  any  point  A,  and  applying 
A  0  six  times  as  a  chord  to  the  circum- 
ference, we  are  brought  round  to  the 
point  of  beginning, 
AB  CD  EF. 


figure 


and  the  inscribed 

thus    formed,   is   a   regular   hex- 
agon. 

357.  Cor.  2.  By  joining  the  alternate  angles  of  the  in- 
scribed regular  hexagon  by  the  straight  lines  A  C,  C  E, 
E  A,  the  figure  ACE,  thus  inscribed  in  the  circle,  will  be 
an  equilateral  triangle,  since  its  sides  subtend  equal  arcs, 
ABC,  CDE,  EFA,  on  the  circumference  (Prop.  III. 
Bk.  III.). 

358.  Cor.  3.  Join  0  A,  0  C,  and  the  figure  A  B  C  0  is 
a  rhombus,  for  each  side  is  equal  to  the  radius.  Hence, 
the  sum  of  the  squares  of  the  diagonals  A  C,  0  B  is  equiv- 
alent to  the  sum  of  the  squares  of  the  sides  (Prop.  XV. 
Bk.  IV.)  ;  or  to  four  times  the  square  of  the  radius 
0  B ;  that  is,  A  C^  -f  OB^  is  equivalent  to  4  AB^  or 
4  6  B^  ;  and  taking  away  0  B"  from  both,  there  remains 
A  C^  equivalent  to  3  0  B  ;  hence 

AC^0B'::3:1,     or     AC  :  0  B  :  :  Vs  :  1 ; 

hence,  the  side  of  the  inscribed  equilateral  triangle  is  to 
the  radius  as  the  square  root  of  3  is  to  unit?/. 


Proposition  VI.  —  Problem. 

359.     To    inscribe    a    regular    decagon   in   a    given 
circle. 

Divide  tlie  radius,  0  A,  of  the  given  circle,  in  extreme 
and  mean  ratio,  at  the  point  M  (Prob.  XXVII.  Bk.  V.). 


148  ELEMENTS  OF  GEOMETRY. 

Take  the  chord  AB  equal  to  OM, 
and  AB  will  be  tlie  side  of  a  regular 
decagon  inscribed  in  the  circle.  For 
we  have  by  construction, 

A  0  :  0  M  :  :  0  M  :  A  M  ; 

or,  since  A  B  is  equal  to  0  M, 
A  0  :  A  B  :  :  A  B  :  A  M. 

Draw  MB  and  BO;  and  the  triangles  ABO,  AMB 
have  a  common  angle,  A,  included  between  proportional 
sides  ;  hence  the  two  triangles  are  similar  (Prop.  XXIV.. 
Bk.  IV.).  Now,  the  triangle  OAB  being  isosceles,  AMB 
must  also  be  isosceles,  and  A  B  is  equal  to  B  M  ;  but  A  B 
is  equal  to  0  M,  consequently  M  B  is  equal  to  M  0 ;  hence 
the  triangle  M  B  0  is  isosceles. 

Again,  the  angle  AMB,  being  exterior  to  the  isosceles 
triangle  BMO,  is  double  the  interior  angle  0  (Prop. 
XXVII.  Bk.  L).  But  the  angle  AMB  is  equal  to  the 
angle  M  A  B  ;  hence  the  triangle  0  A  B  is  such,  that  each 
of  the  angles  at  the  base,  OAB,  OB  A,  is  double  the  angle 
0,  at  its  vertex.  Hence  the  three  angles  of  tlie  triangle 
are  together  equal  to  five  times  the  angle  0,  which  conse- 
quently is  a  fifth  part  of  two  right  angles,  or  the  tenth 
part  of  four  right  angles  ;  therefore  the  arc  A  B  is  the 
tenth  part  of  the  circumference,  and  the  chord  A  B  is  the 
side  of  an  inscribed  regular  decagon. 

360.  Cor.  1.  By  joining  the  vertices  of  the  alternate 
angles  A,  C,  &g.  of  the  regular  decagon,  a  regular  penta- 
gon may  be  inscribed.  Hence,  the  chord  A  C  is  the  side 
of  an  inscribed  regular  pentagon. 

361.  Cor.  2.  A  B  being  the  side  of  the  inscribed  regu- 
lar decagon,  let  AL  be  tlie  side  of  an  inscribed  regular 
hexagon  (Prop.  V.  Cor.  1).  Join  BL;  then  BL  will  be 
the  side  of  an  inscribed  regular  pentedecagon,  or  regular 
polygon  of  fifteen  sides.  For  AB  cuts  off  an  arc  equal  to 
a  tenth  part  of  the  circumference  ;  and  A  L  subtends  an 


BOOK  VI.  149 

arc  equal  to  a  sixth  of  the  circumference  ;  therefore  B  L, 
tlie  difference  of  these  arcs,  is  a  fifteenth  part  of  the  cir- 
cumference ;  and  since  equal  arcs  are  subtended  by  equal 
chords,  it  follows  that  the  chord  B  L  may  be  a])|)licd  ex- 
actly fifteen  times  around  the  circumference,  thus  forming 
a  regular  pentedecagon. 

362.  Scholium.  If  the  arcs  subtended  by  the  sides  of 
any  inscribed  regular  polygon  be  severally  bisected,  the 
chords  of  those  semi-arcs  will  form  another  inscribed 
polygon  of  double  the  number  of  sides.  Thus,  from 
having  an  inscribed  square,  there  may  be  inscribed  in  suc- 
cession polygons  of  8,  10,  32,  64,  <fec.  sides  ;  from  the 
hexagon  may  be  formed  polygons  of  12,  24,  48,  96,  <fec. 
sides  ;  from  the  decagon,  polygons  of  20,  40,  80,  <fec.  sides  ; 
and  from  the  pentedecagon,  polygons  of  30,  60,  120,  &c. 
sides. 

Note.  —  For  a  long  time  the  polygons  above  noticed  were  supposed 
to  include  all  that  could  be  inscribed  in  a  ciix:le.  In  the  year  1801,  M. 
Gauss,  of  Gottingen,  made  known  the  curious  discovert'  that  the  circum- 
ference of  a  circle  could  be  divided  into  any  number  of  equal  parts 
capable  of  being  expressed  by  the  formula  2"  -[-  1,  provided  it  be  a 
prime  number.  Now,  the  number  3  is  the  simplest  of  this  kind,  it  being 
the  value  of  the  above  formula  when  the  exponent  n  is  1 ;  the  next 
prime  number  is  5,  and  this  is  contained  in  the  formula.  But  the  poly- 
gons of  3  and  of  5  sides  have  already  been  inscribed.  The  next  prime 
number  expressed  by  the  fonnula  is  1 7,  so  that  it  is  possible  to  inscribe 
a  regular  polygon  of  17  sides  in  a  circle.  The  investigations,  however, 
-which  establish  this  geometrical  fact  involve  considerations  of  a  nature 
that  do  not  enter  into  the  elements  of  Geometry. 

Proposition  YII.  —  Problem. 

363.  A  regular  inscribed  polygon  being  given,  to  cir- 
cumscribe a  similar  polygon  about  the  same  circle. 

Let  ABCDEF  be  a  regular  polygon  inscribed  in  a 
circle  whose  centre  is  0. 

Through  M,  the  middle  point  of  the  arc  A  B,  draw  the 
tangent,  G II ;  also  draw  tangents  at  the  middle  points  of 

13* 


150  ELEMENTS  OP  GEOMETRY. 

tlie  arcs  B  C,  C  D,  <fec. ;  these  tan- 
gents are  parallel  to  the  chords 
AB,  BC,  CD,  &c.  (Prop.  XL 
Bk.  IIL,  and  Prop.  YI.  Cor.  1, 
Bk.  III.),  and  by  their  intersec- 
tions form  the  regular  circum- 
scribed polygon  GHI,  &c.  similar 
to  the  one  inscribed. 

Since  M  is  the  middle  point  of  the  arc  A  B,  and  N  tlie 
middle  point  of  the  equal  arc  B  C,  the  arcs  B  M,  B  N  are 
halves  of  equal  arcs,  and  therefore  are  equal ;  that  is,  the 
vertex,  B,  of  the  inscribed  polygon  is  at  the  middle  point 
of  the  arc  MN.  Draw  OH ;  the  line  OH  will  pass  through 
the  point  B.  For,  the  right-angled  triangles  OMH,  ONH, 
having  the  common  hypothenuse  0  H,  and  the  side  0  M 
equal  to  ON,  must  be  equal  (Prop.  XIX.  Bk.  I.),  and 
consequently  the  angle  M  0  H  is  equal  to  HON,  where- 
fore the  line  0  H  passes  through  the  middle  point,  B,  of 
the  arc  M  N.  In  like  manner,  it  may  be  shown  that  the 
line  01  passes  through  the  middle  point,  C,  of  the  arc 
N  P  ;  and  so  with  the  other  vertices. 

Since  O  H  is  parallel  to  A  B,  and  II I  to  B  C,  the  angle 
G  II I  is  equal  to  the  angle  ABC  (Prop.  XXVI.  Bk.  I.)  ; 
in  like  manner,  H  I  K  is  equal  to  B  C  D  ;  and  so  with  the 
other  angles  ;  hence,  the  angles  of  the  circumscribed  poly- 
gon are  equal  to  those  of  the  inscribed  polygon.  And, 
further,  by  reason  of  these  same  parallels,  we  have 

GH:AB::OH:OB,  and  H  I  :  B  C  :  :  0  H  :  OB ; 

therefore  (Prop.  X.  Bk.  II.), 

G  H  :  A  B  :  :  H  I  :  B  C. 

But  A  B  is  equal  to  B  C,  therefore  G  H  is  equal  to  II  I. 
For  the  same  reason,  H  I  is  equal  to  I K,  <fec. ;  conse- 
quently, the  sides  of  the  circumscribed  polygon  are  all 
equal ;  hence  this  polygon  is  regular,  and  similar  to  the 
inscribed  one. 


BOOK   VI.  151 

364.  Cor.  1.  Converselj,  if  the  circumscribed  polygon 
G  H  I  K,  <fcc.  is  given,  and  it  is  required,  by  means  of  it, 
to  construct  a  similar  inscribed  polygon,  draw  the  straight 
lines  0  G,  OH,  <fcc.  from  the  vertices  of  the  angles  G,  H, 
I,  <fec.  of  the  given  polygon  to  the  centre  ;  tlie  lines  will 
meet  the  circumference  in  tlie  points  A,  B,  C,  <fec.  Join 
these  points  by  the  chords  A  B,  B  C,  <fec.,  which  will  form 
the  inscribed  polygon.  Or  simply  join  the  points  of  con- 
tact, M,  N,  P,  <fec.,  by  chords,  MN,  N  P,  <fcc.,  which  like- 
wise would  form  an  inscribed  polygon  similar  to  the  cir- 
cumscribed one. 

365.  Cor.  2.  Hence,  we  may  circumscribe  about  a  cir- 
cle any  regular  polygon  similar  to  an  inscribed  one,  and 
conversely. 

366.  Cor.  3.  It  has  been  shown  that  N  H  and  H  M  are 
equal ;  therefore  the  sum  of  N  H  and  H  M,  which  is  equal 
to  the  sum  of  H  M  and  M  G,  is  equal  to  H  G,  one  of  the 
equal  sides  of  the  polygon. 

367.  Scholium.  From  having  a  circumscribed  regular 
polygon,  another  having  double  the  number  of  sides  may 
be  readily  constructed,  by  drawing  tangents  to  the  points 
of  bisection  of  tlie  arcs,  intercepted  by  the  sides  of  the  pro- 
posed polygon,  limiting  these  tangents  by  those  sides.  In 
like  manner  other  circumscribed  polygons  may  be  formed  ; 
l)ut  it  is  plain  that  each  of  the  polygons  so  formed  will  be 
less  than  the  preceding  polygon,  being  entirely  compre- 
hended in  it. 

Proposition  YIH. — Theorem. 

368.  Tlie  area  of  a  regular  poly ^^on  is  equivalent  to  the 
product  of  its  perimeter  hy  half  of  the  radius  of  the  in- 
scribed circle. 

Let  A  B  C  B  E  F  be  a  regular  polygon,  and  0  tlie 
centre  of  the  inscribed  circle. 

From  0  let  the  straight  lines  0  A,  0  B,  <fec.  be  drawn  to 


152 


ELEMENTS   OP   GEOMETRY. 


E 

D 

^ 

0 

^ 

y 

•■';■■.'" 

■"]■"/ 

/    •    '■•• 

■•-•.. 

// 

/ 

1 

\  y 

/n 

^^ 

^-.7 

A 

M 

B 

the  vertices  of  all  the  angles  of 
the  polygon,  and  the  polygon  will 
be  divided  into  as  many  equal 
triangles  as  it  has  sides  ;  and  let 
the  radii  0 M,  ON,  &c.  of  the  in- 
scribed circle  be  drawn  to  the 
centres  of  the  sides  of  the  polygon, 
or  to  the  points  of  tangency  M, 
N,  <fec.,  and  these  radii  are  perpendicular  to  the  sides  re- 
spectively (Prop.  XI.  Bk.  III.)  ;  therefore  the  radius  of 
the  circle  is  equal  to  the  altitude  of  the  several  triangles. 

Now,  the  triangle  A  0  B  is  measured  by  the  product  of 
AB  by  half  of  0  M  (Prop.  YI.  Bk.  IV.)  ;  the  triangle 
O  B  C  by  the  product  of  B  C  by  half  of  0  N.  But  0  M  is 
equal  to  0  N  ;  hence  the  two  triangles  taken  together  are 
measured  by  the  sum  of  A  B  and  B  C  by  half  of  0  M.  In 
like  manner  the  measure  of  the  other  triangles  may  be 
found  ;  hence,  the  sum  of  all  the  triangles,  or  the  whole 
polygon,  is  equal  to  the  sum  of  the  bases  AB,  B  C,  ttc, 
or  the  perimeter  of  tho  polygon,  multiplied  by  half  of  OM, 
or  half  the  radius  of  the  inscribed  circle. 


Proposition  IX.  —  Theorem. 

369.  The  perimeters  of  two  reg-vlar  polygons^  having 
the  same  number  of  sides,  are  to  each  other  as  the  radii 
of  the  circumscribed  circles,  and,  also,  as  the  radii  of  the 
inscribed  circles  ;  and  their  areas  are  to  each  other  as  the 
squares  of  those  radii. 

Let  A  B  be  a  side  of  one  polygon, 
O  the  centre,  and  consequently  0  A 
the  radius  of  the  circumscribed  cir- 
cle, and  0  M,  perpendicular  to  A  B, 
the  radius  of  the  inscribed  circle. 
Let  G  H  be  a  side  of  the  other  poly- 
gon, C  the  centre,  C  G  and  C  N  the  A  B  G 
radii  of  the  circumscribed  and  the  inscribed  circles. 


The  perimeters  of  tlie  two  polygons  arc  to-e*eli  otIi'Sr' 
the  sides  AB  and  G  H  (Prop.  XXXI.  Bk.  iyr)rf)til^ 
angles  A  and  G  are  equal,  being  eacli  luilf  oi'  tlio  angle  of 
the  polygon  ;  so  also  are  the  angles  B  and  H  ;  hence, 
drawing  OB  and  OH,  the  isosceles  triangles  ABO,  GHC 
are  similar,  as  are  likewise  the  right-angled  triangles 
A  M  0,  G  N  C  ;  hence 

A  B  :  G  H  :  :  A  0  :  G  C  :  :  M  0  :  N  C. 

Hence,  the  perimeters  of  the  polygons  are  to  each  other  as 
the  radii  A 0,  GO  of  tlie  circumscribed  circles,  and,  also, 
as  tlie  radii  M  0,  N  C  of  the  inscribed  circles. 

The  surfaces  of  these  polygons  are  to  each  other  as  the 
squares  of  the  homologous  sides  A  B,  G  H  (Prop.  XXXI. 
Bk.  IV.)  ;  tliey  are  therefore  to  each  other  as  the  squares 
of  AO,  GO,  the  I'adii  of  the  circumscribed  circles,  or  as 
the  squares  of  0  M,  ON,  the  radii  of  the  inscribed  circles. 

Proposition  X.  —  Pboblem. 

370.  The  surface  of  a  reg'iilar  inscribed  poii/g'on,  and 
that  of  a  similar  circumscribed  pol/jg-on,  bei/i^  g-iven  ;  to 
find  the  surfaces  of  regular  inscribed  and  circumscribed 
polygons  having  double  the  number  of  sides. 

Let  AB  be  a  side  of  the  given      ^      -r.     ,,      ^     „ 

M    J       1  T^Ti  iTix        EPMQF 

inscribed  polygon  ;  l!j  F,  parallel  to 

AB,   a   side  of  the   circumscribed 

polygon,  and  C  the  centre  of  tlie 

circle.     Draw  the  chord  AM,  and 

the  tangents  A  P,  B  Q  ;  then  A  M 

will  be  a  side  of  the  inscribed  poly-  v 

gon,  having   twice   the   number  of  ^ 

sides  ;  and  P  Q,  the  double  of  P  M,  will  be  a  side  of  the 

similar  circumscribed  polygon. 

Let  A,  then,  be  the  surface  of  the  inscribed  polygon 

whose  side  is  AB,  B  tliat  of  tlie  similar  circumscribed 

polygon ;  A'  the  surface  of  the  polygon  whose  side  is  A  ^M, 


154  ELEMENTS  OP  GEOMETRY. 

B'  that  of  the  similar  circumscribed      ^      ^     ^.      ^     -,, 
1  A         A    Ty  •  E       P      M       Q     F 

polygon  :  A  and  B  are  given  ;  we       . — -7— 

have  to  find  A'  and  B'.  4^^^ 

First.  The  triangles  ACD,ACM,  \\     f      / 

whose  common  vertex  is  A,  are  to 
each  other  as  their  bases  CD,  CM 
(Prop.  VI.  Cor.,  Bk.  lY.)  ;  they  are 
likewise  as  the  polygons  A  and  A' ; 

hence 

A  :  A'  :  :  C  D  :  C  M. 

Again,  the  triangles  CAM,  C  M  E,  whose  common  vertex 
is  M,  are  to  each  other  as  their  bases  C  A,  C  E  ;  they  are 
likewise  to  each  other  as  the  polygons  A'  and  B ;  hence 

A'  :  B  :  :  C  A  :  C  E. 

But,  since  A  D  and  M  E  are  parallel,  we  have, 

CD  :  CM:  :  CA:  CE; 
hence 

A  :  A'  :  :  A'  :  B ; 

hence,  the  polygon  A!  is  a  mean  proportional  between  tliei 
two  given  polygons. 

Secondly.  The  altitude  C  M  being  common,  the  tri- 
angle CPM  is  to  the  triangle  C  PE  as  PM  is  to  PE ; 
but  since  CP  bisects  the  angle  MCE,  we  have  (Prop. 
XIX.  Bk.  IV.), 

PM:PE::CM:CE::CD:CA::A:A'; 

hence 

CPM:  CPE:  :  A:  A'; 

and,  consequently, 

C  P  M  :  C  P  M  +  C  P  E  or  C  M  E  :  :  A  :  A  +  A'. 
But  CMPAor2  CMP  and  C  M  E  are  to  each  other  as 
the  polygons  B'  and  B  ;  hence 

B'  :  B  :  :  2  A:  A  + A'; 

which  gives 

J\i  —  2  Ax  B 
-^  —  A  +  A'  ' 


BOOK  VI.  155 

or,  the  polygon  B'  is  equal  to  the  qvotient  of  ttoice  the 
product  of  the  given  polygons  divided  by  the  sum  of  the 
inscribed  polygons. 

Thus,  by  means  of  the  polygons  A  and  B,  it  is  easy  to 
find  the  polygons  A'  and  B',  whicli  have  double  the  num- 
ber of  sides. 

Proposition  XI.  —  Theorem. 

371.  A  circle  being  given ^  tivo  similar  polygons  can 
alvmys  be  formed^  the  one  circumscribed  about  the  circle^ 
the  other  inscribed  in  it^  which  shall  differ  from  each  other 
by  less  than  any  assignable  surface. 

Let  Q  be  the  side  of  a  square  I 

less  than  the  given  surface.  H,^^^^^^^^:^^  |r 

Bisect  A  C,  a  fourth  part  of  ^^\.  3)U 

the  circumference,  and  then  bi-  /jsr-\.\       "^••^..    \V 

sect  the  half  of  this  fourth,  and  ^  u  a  "o  W/^'^ 
so  proceed  until  an  arc  is  found  \  o  // 

whose  chord  AB  is  less  tliau  ^^>J        Z-^ 

Q.     As  this  arc  must  be  an  ex-  ^^^"rtp^^^"^ 

act  part  of  the  circumference,  if  we  apply  the  chords  AB, 
B  C,  <fec.,  each  equal  to  AB,  the  last  will  terminate  at  A, 
and  there  will  be  inscribed  in  the  circle  a  regular  polygon, 
ABODE,  &c.  Next  describe  about  the  circle  a  similar 
polygon,  G  H I K  L,  &c.  (Prop.  VII.)  ;  and  the  difference 
of  tliese  two  polygons  will  be  less  than  the  square  of  Q. 

Find  the  centre,  0  ;  from  the  points  G  and  H  draw  the 
straight  lines  GO,  HO,  and  they  will  pass  through  the 
points  A  and  B  (Prop.  VII.).  Draw  also  OM  to  the  point 
of  tangency,  M ;  and  it  will  bisect  A  B  hi  N,  and  be  per- 
pendicular to  it  (Prop.  VI.  Cor.  1,  Bk.  III.)-  Produce 
A  0  to  E,  and  draw  B  E. 

Let  P  represent  the  circumscribed  polygon,  and  p  the 
inscribed  polygon.  Thou,  since  these  polygons  are  simi- 
lar, they  are  as  the  squares  of  the  homologous  sides  G  H, 


1.G6  ELEMENTS   OF   GEOMETRY. 

A  B  (Prop.  XXXI.  Bk.  lY.)  ;  but  the  triangles  G  0  H, 
A  O  B  are  similar  (Prop.  XXIV.  Bk.  IV.)  ;  hence  tliey 
are  to  each  otlier  as  the  squares  of  the  homologous  sides 
OG  and  0  A  (Prop.  XXIX.  Bk.  IV.)  ;  therefore 

P  :  ;?  :  :  OG^  OA'  or  0  III 
Again,  the  triangles  O  G  M,  E  A  B,  having  their  sides 
respectively  parallel,  are  similar ;  therefore 

F:p::  OW  :  OM^  :  :  A  E'  :  Blf ; 
and,  hy  division, 

P  :  P  —p  :  :  A  E^  :  AE^  — ¥B^  or  AB"l 
But  P  is  less  than  tlie  square  described  on  the  diameter 
AE;  therefore  P — p  is  less  than  tlie  square  described 
on  AB,  that  is,  less  than  the  given  square  Q.  Hence, 
tlie  difference  between  the  circumscribed  and  inscribed 
polygons  may  always  be  made  less  than  any  given  sur- 
face. 

372.  Cor.  Since  the  circle  is  obviously  greater  than  any 
inscribed  polygon,  and  less  than  any  circumscribed  one,  it 
follows  that  a  po/j/g-on  may  be  inscribed  or  circumscribed^ 
which  ivill  differ  from  the  circle  by  less  than  any  assign- 
able magnitude. 

Proposition  XII.  — Problem. 

373.  To  find  the  approximate  area  of  a  circle  vjhose 
radius  is  unity. 

Let  the  radius  of  the  circle  ])o  1,  and  let  tlie  first  in- 
scribed and  circumscribed  polygons  be  squares  ;  the  side 
of  the  inscribed  square  will  be  V:^  (Prop.  IV.  Cor.),  and 
that  of  the  circumscribed  square  will  be  equal  to  tlie  di- 
ameter 2.  Hence  the  surface  of  the  inscribed  square  is 
2,  and  that  of  the  circumscribed  square  is  4.  Let,  there- 
fore A  =  2,  and  B  =  4.  Now  it  has  been  proved,  in 
Proposition  X.,  that  tlie  surface  of  the  inscribed  octagon, 
or,  as  it  has  been  represented,  A',  is  a  mean  proportional 


BOOK   VI.  157 

between  the  two  squares  A  and  B,  so  that  A'  =  v'  8  =. 
2.8284271;  and  it  has  also  been  proved,  in  the  same  prop- 
osition, that  the  circumscribed  octagon,  represented  by  B', 

=  ^A^^l  so  that  B'  =  -i^  =  3.3137085.      The 

inscribed  and  the  circumscribed  octagons  being  thus 
determined,  we  can  easily,  by  means  of  them,  determine 
the  polygons  having  twice  the  number  of  sides.  We  have 
only  in  this  case  to  put  A  =  2.8284271,  B  =  3.3137085  ; 
and  we   shall   find  A'  =  V  A  X  B  =  3.0G14674,   and 

B'  =   2^iX^  =  3.1825979. 
A-f  A' 

In  like  manner  may  be  determined  the  area  of  polygons 
of  sixteen  sides,  and  thence  the  area  of  polygons  of  thirty- 
two  sides,  and  so  on  till  we  arrive  at  an  inscribed  and 
a  circumscribed  polygon  differing  so  little  from  each  other, 
and  consequently  fi'om  the  circle,  that  the  difference  shall 
be  less  than  any  assignable  magnitude  (Prop.  XI-  Cor.). 

The  subjoined  table  exhibits  the  area,  or  numerical  ex- 
pression for  the  surface,  of  these  polygons,  carried  on  till 
they  agree  as  far  as  the  seventh  place  of  decimals. 

Number  of  sides.  Inscribed  Polvsjons.  Circumscribed  Polygons. 

4  .  .  .  .  2.0000000  ....  4.0000000 

8  .  .  .  .  2.8284271  ....  3.3137085 

IG  .  .  .  .  3.0614674  ....  3.1825979 

32  ...  .  3.1214451  ....  3.1517249 

64  ...  .  3.1365485  ....  3.1441148 

128  ...  .  3.1403311  ....  3.1422236 

256  ....  3.1412772  ....  3.1417504 

512  ...  .  3.1415138  ....  3.1416321 

1024  ....  3.1415729  ....  3.1416025 

2048  ....  3.1415877  ....  3.1415951 

4096  ....  3.1415914  ....  3.1415933 

8192  .....  3.1415923  ....  3.1415928 

16384  .  .  ."  .  3.1415925  ....  3.1415927 

32768  ....  3.1415926  ....  3.1415926 

14 


158  ELEMENTS  OF  GEOMETRY. 

It  appears,  therefore,  that  the  inscribed  and  circum- 
scribed polygons  of  32768  sides  differ  so  little  from  each 
other  that  the  numerical  value  of  each,  as  far  as  seven 
places  of  decimals,  is  absolutely  the  same  ;  as  the  circle  is 
between  the  two,  it  cannot,  strictly  speaking,  differ  from 
either  so  much  as  they  do  from  each  other ;  so  that  the 
number  3.1415926  expresses  the  area  of  a  circle  whose 
radius  is  1,  correctly,  as  far  as  seven  places  of  decimals. 

Some  doubt  may  exist,  perhaps,  about  the  last  decimal 
figure,  owing  to  errors  proceeding  from  the  parts  omitted ; 
but  the  calculation  has  been  carried  on  with  an  additional 
figure,  that  the  final  result  here  given  might  be  absolutely 
correct  even  to  the  last  decimal  place. 

374.  Cor.  Since  the  inscribed  and  circumscribed  poly- 
gons are  regular,  and  have  the  same  number  of  sides,  they' 
are  similar  (Prop.  I.)  ;  therefore,  by  increasing  the  num- 
ber of  the  sides,  the  corresponding  polygons  formed  will 
approach  to  an  equality  with  the  circle.  Now  if,  by  con- 
tinual bisections,  the  polygons  formed  shall  have  their 
number  of  sides  indefinitely  great,  each  side  will  become 
indefinitely  small,  and  the  inscribed  and  circumscribed 
polygons  will  ultimately  coincide  with  each  other.  But 
when  they  coincide  with  each  other,  they  must  each  co- 
incide with  the  circle,  since  no  part  of  an  inscribed  poly- 
gon can  be  without  the  circle,  nor  can  any  part  of  a 
circumscribed  one  be  within  it ;  hence,  the  perimeters  of 
the  polygons  must  coincide  with  the  circumference  of 
the  circle^  and  be  equal  to  it. 

375.  Scholium.  Every  circle,  therefore,  may  be  regard- 
ed as  a  polygon  of  an  infinite  number  of  sides. 

Note.  —  This  new  definition  of  the  circle,  if  it  does  not  appear  at 
first  view  to  be  very  strict,  has  at  least  the  advantage  of  introducing 
more  simplicity  and  precision  into  demonstrations.  {Cows  de  Ge'ome- 
trie  Elementaire,  par  Vincent  et  Bourdon.) 


BOOK    VI. 


159 


Proposition  XIII. — Theorem. 


376.    TJ\e  circumferences  of  circles  are  to  each  other  as 
their  radii,  and  their  areas  are  to  each  other  as  the  squares 
of  their  radii. 
Let  C  denote 
tlie  circumfer- 
ence of  one  of 
tlie  circles,  R 
its  radius  OA, 
A  its  area;  and 
let   C   denote 
tlie  circumfer- 
ence of  the  otlier  circle,  r  its  radius  0  B,  A'  its  area ;  tlien 
will 

C  :  C  :  :  R  :  r, 
and 

A  :  A'  :  :  R^  :  r\ 

Inscribe  within  the  given  circles  two  regular  polygons 
of  the  same  number  of  sides  ;  and,  whatever  be  the  num- 
ber of  sides,  the  perimeters  of  the  polygons  will  be  to  each 
other  as  the  radii  0  A  and  OB  (Prop.  IX.).  Now,  con- 
ceive the  arcs  subtending  the  sides  of  the  polygon  to  be 
continually  bisected,  forming  other  inscribed  polygons, 
until  polygons  are  formed  of  an  indefinite  number  of  sides, 
and  therefore  having  perimeters  coinciding  with  the  cir- 
cumference of  the  circumscribed  circles  (Prop.  XII.  Cor.)  ; 
and  we  shall  have 

C  :  C  :  :  R  :  r. 

Again,  the  areas  of  the  inscribed  polygons  are  to  each 
other  as  0  A  to  OB  (Prop.  IX.).  But  when  the  num- 
ber of  sides  of  tlie  polygons  is  indefinitely  increased,  the 
areas  of  the  polygons  become  equal  to  the  areas  of  the 
circles  ;  hence  we  shall  have 

A  :  A'  :  :  R2  :  r^ 


160  ELEMENTS  OF  GEOMETRY. 

377.  Cor.  1.  The  circumferences  of  circles  are  to  each 
other  as  twice  their  radii,  or  as  tlieir  diameters. 

For,  multiplying  the  terms  of  the  second  ratio  in  the 
first  proportion  by  2,  we  have 

C  :  C  :  :  2  R  :  2  r. 

378.  Cor.  2.  The  areas  of  circles  are  to  each  other  as 
the  squares  of  their  diameters. 

For,  multiplying  the  second  ratio  of  the  second  propor- 
tion by  4,  or  2  squared,  we  have 

A:  A':  :4R^•4r^ 

Proposition  XI Y.  —  Theorem. 

379.  Similar  arcs  are  to  each  other  as  their  radii;  and 
similar  sectors  are  to  each  other  as  the  squares  of  their 
radii. 

Let  A  B,  D  E  be  similar 
arcs  ;  A  C  B,  DOE,  similar 
sectors ;  and  denote  the  radii 
C  A  and  OD  by  R  and  r ;  then 
will 

A  B  :  D  E  :  :  R  :  r, 

and  A  C  B  :  D  0  E  :  :  R2 :  r^. 

For,  since  the  arcs  are  similar,  the  angle  C  is  equal  to 
the  angle  0  (Art.  213).  But  the  angle  C  is  to  four  right 
angles  as  the  arc  AB  is  to  the  whole  circumference  de- 
scribed with  the  radius  C  A  (Prop.  XVII.  Sch.  2,  Bk.  III.)  ; 
and  the  angle  0  is  to  four  right  angles  as  the  arc  D  E 
is  to  the  circumference  described  with  the  radius  OD. 
Hence,  the  arcs  A  B,  D  E  are  to  each  other  as  the  circum- 
ferences of  which  they  form  a  part.  But  these  circumfer- 
ences are  to  each  other  as  their  radii,  C  A,  0  D  (Prop. 
XIII.)  ;  therefore 

^rc  A  B  :  ^rc  D  E  :  :  R  :  r. 

By  like  reasoning,  the  sectors  A  C  B,  DOE  are  to  each 


BOOK   VI. 


IGl 


otlicr  as  tlie  whole  circles  of  which  they  are  a  part ;  and 
these  are  as  the  squares  of  tlieir  radii  (Prop.  XIII.)  ; 
therefore 

Sector  A  C  B  :  Sector  D  0  E  :  :  R^  :  r^. 


Proposition  XY.  —  Theorem. 

380.  The  area  of  a  circle  is  equal  to  the  product  of  the 
tircumference  by  half  the  radius. 

Let  C  denote  the  circumference  of 
the  circle,  whose  centre  is  0,  R  its 
radius  0  A,  and  A  its  area ;  then  will 
A  =  C  X  ^  R. 

For,  inscribe  in  the  circle  any  reg- 
ular polygon,  and  from  the  centre 
draw  OP  perpendicular  to  one  of  the 
sides.  The  area  of  the  polygon,  whatever  be  the  number 
of  sides,  will  be  equal  to  its  perimeter  multiplied  by  half 
of  OP  (Prop.  VIII.).  Conceive  the  arcs  subtending  the 
sides  of  the  polygon  to  be  continually  bisected,  until  a 
polygon  is  formed  having  an  indefinite  number  of  sides  ; 
its  perimeter  will  be  equal  to  the  circumference  of  tlie 
circle  (Prop.  XII.  Cor.),  and  0  P  be  equal  to  the  radius 
0  A  ;  therefore  the  area  of  the  polygon  is  equal  to  that  of 
the  circle  ;  hence 

A  =  C  X  i-  R. 

381.  Cor.  1.  The  area  of  a  sector 
is  equal  to  the  product  of  its  arc  by 
half  of  its  radius. 

For,  let  C  denote  the  circumference 
of  the  circle  of  which  the  sector  DOE 
is  a  part,  R  its  radius  0  D,  and  A  its 
area  ;  tlien  we  sliall  have  (Prop. 
XYII.  Sch.  2,  Bk.  III.), 

Sector  D  0  E  :  A  :  :  Arc  D  E  :  C ; 


162  ELEMENTS  OF  GEOMETRY. 

hence,  since  equimultiples  of  two  magnitudes  have  the 
same  ratio  as  tlie  magnitudes  themselves  (Prop.  IX.  Bk. 
IL), 

Sector  D  0  E  :  A  :  :  ^rc  D  E  X  i^  R  :  C  X  i  R. 

But  A,  or  the  area  of  the  whole  circle,  is  equal  to 
C  X  2"  R ;  hence,  the  area  of  the  sector  D  0  E  is  equal 
to  the  arc  D  E  X  ^  R. 

382.  Cor.  2.  Let  the  circumference  of  the  circle  whose 
diameter  is  unity  be  denoted  by  n  (which  is  called  pi), 
the  radius  by  R,  and  the  diameter  by  D  ;  and  the  circum- 
ference of  any  other  circle  by  C,  and  its  area  by  A. 
Then,  since  circumferences  are  to  each  other  as  their 
diameters  (Prop.  XIII.  Cor.  1),  we  shall  have, 

C  :  D  :  :  tt:  1; 
therefore 

C  =  DX^=2RX^. 

Multiplying  both  numbers  of  this  equation  by  ^  R,  we  have 

C  X  i-  R  =  R'  X  ^,    or    A  =  R^  X  ^ ; 
tliat  is,  the  area  of  a  circle  is  equal  to  the  product  of  the 
square  of  its  radius  by  the  constant  number  n. 

383.  Cor.  3.  The  circumference  of  every  circle  is  equal 
to  the  product  of  its  diameter,  or  twice  its  radius,  by  the 
constant  number  n. 

384.  Cor.  4.  The  constant  number  n  denotes  the  ratio 
of  the  circumference  of  any  circle  to  its  diameter ;  for 

C 

D  =  "- 

385.  Scholium  1.  The  exact  numerical  value  of  the 
ratio  denoted  by  n  can  be  only  approximately  expressed. 
The  approximate  value  found  by  Proposition  XII.  is 
3.1415926  ;  but,  for  most  practical  purposes,  it  is  suf- 
ficiently accurate  to  take  n  =  3.1416.  The  symbol  n  is 
the  first  letter  of  tlie  Greek  word  ireplfMerpop,  pcrimetron, 
which  signifies  circumference. 


BOOK   VI.  163 

386.  Scholium  2.  The  Quadrature  of  the  Circle  is 
the  problem  "vvhich  requires  tiie  finding  of  a  square  equiv- 
alent in  area  to  a  circle  having  a  given  radius.  Now,  it 
has  just  been  proved  that  a  circle  is  equivalent  to  the  rec- 
tangle contained  by  its  circumference  and  half  its  radius  ; 
and  this  rectangle  may  be  changed  into  a  square,  by  find- 
ing a  mean  proportional  between  its  length  and  its  breadth 
(Prob.  XXVI.  Bk.  Y.).  To  square  the  circle,  therefore, 
is  to  find  the  circumference  when  the  radius  is  given ;  and 
for  effecting  this,  it  is  enough  to  know  the  ratio  of  the  cir- 
cumference to  its  radius,  or  its  diameter. 

But  this  ratio  has  never  been  determined  except  approx- 
imately ;  but  the  approximation  has  been  carried  so  far, 
that  a  knowledge  of  the  exact  ratio  would  afford  no  real 
advantage  whatever  beyond  that  of  the  approximate  ratio. 
Professor  Rutherford  extended  the  approximation  to  208 
places  of  decimals,  and  Dr.  Clausen  to  250  places.  The 
value  of  71,  as  developed  to  208  places  of  decimals,  is 
3 . 14159265358979323846264338327950288419716939937 
5105820974944592307816406286208998628034825342717 
0G79821480865132823066470938446095505822317253594 
0812848473781392038633830215747399600825931259129 
40183280651744. 

Such  an  approximation  is  evidently  equivalent  to  per- 
fect correctness  ;  the  root  of  an  imperfect  power  is  iu 
no  case  more  accurately  known. 

Proposition  XYI.  —  Problem. 

387.  To  divide  a  circle  into  any  number  of  equal  parts 
by  means  of  concentric  circles. 

Let  it  be  proposed  to  divide  the  circle,  whose  centre  is 
0,  into  a  certain  number  of  equal  parts,  —  three  for  in- 
stance,—  by  means  of  concentric  circles. 

Draw  the  radius  AO;  divide  AO  into  three  equal  parts, 
A  B,  B  C,  C  0.     Upon  A  0  describe  a  semi-circumference, 


164 


ELEMENTS   OF   GEOMETRY. 


and  draw  the  perpendiculars,  B  E, 
C  D,  meeting  that  semi-circumfer- 
ence in  the  points  E,  D.  Join 
0  E,  0  D,  and  with  these  Hnes  as 
radii  from  the  centre,  0,  describe 
circles  ;  these  circles  will  divide 
the  given  circle  into  the  required 
number  of  equal  parts. 

For  join  A  E,  AD;  then  the  angle  ADO,  being  in  a 
semicircle,  is  a  right  angle  (Prop.  XVIII.  Cor.  2,  Bk. 
III.)  ;  hence  the  triangles  D  A  0,  D  C  0  are  similar,  and 
consequently  are  to  each  other  as  the  squares  of  their 
homologous  sides  ;  that  is," 


but 


hence 


DAO:DCO:  :  0  A^-  0  D^ 

DAO:  DCO:  :  OA:  00; 

OA':  OD^:  :  OA:  00; 


consequently,  since  circles  are  to  each  other  as  the  squares 
of  their  radii  (Prop.  XIII.),  it  follows  that  tlie  circle 
whose  radius  is  OA,  is  to  that  whose  radius  is  OD,  as  OA 
to  0  C ;  that  is  to  say,  the  latter  is  one  third  of  tlie  former. 
In  the  same  manner,  by  means  of  the  right-angled  tri- 
angles E  A  0,  E  B  0,  it  may  be  proved  that  tlie  circle 
whose  radius  is  0  E,  is  two  thirds  that  whose  radius  is 
0  A.  Hence,  the  smaller  circle  and  the  two  surrounding 
annular  spaces  are  all  equal. 

Note.  —  This  useful  problem  was  first  solved  by  Dr.  Hutton,  the 
justly  distinguished  English  mathematician. 


BOOK   VII 


PLANES.  —  DIEDRAL  AND  POLYEDRAL   ANGLES. 


DEFINITIONS. 


388.  A  STRAIGHT  line  is  per- 
pendiciilar  to  a  plane ^  wlieii  it  is 
perpendicular  to  every  straight 
line  which  it  meets  in  that  plane. 

Conversely,  the  plane,  in  the 
same  case,  is  perpendicular  to  the  line. 

The  foot  of  tlie  perpendicular  is  the  point  in  which  it 
meets  the  plane. 

Thus  the  straight  line  A  B  is  perpendicular  to  the  plane 
M  N ;  the  plane  M  N  is  perpendicular  to  the  straight  line 
A  B  ;  and  B  is  the  foot  of  the  perpendicular  A  B. 

389.  A  line  is  parallel  to  a  plane  when  it  cannot  meet 
the  plane,  however  far  both  of  them  may  be  produced. 

Conversely,  the  plane,  in  the  same  case,  is  parallel  to 
the  line. 

390.  Two  planes  are  parallel  to  each  other,  when  they 
cannot  meet,  however  far  both  of  them  may  be  produced. 

391.  A  PiEDRAL  Angle  is  an 
angle  formed  by  the  intersec- 
tion of  two  planes,  and  is  meas- 
ured by  the  inclination  of  two 
straight  lines  drawn  from  any 
point  in  the  line  of  intersection, 
perpendicular  to  tliat  line,  one 
being  drawn  in  each  plane. 


166 


ELEMENTS    OF    GEOMETRY. 


The  line  of  common  section 
is  called  tlie  edge^  and  the  two 
planes  are  called  the  faces,  of 
the  diedral  angle. 

Thus  the  two  planes  ABM, 
A  B  N,  whose  line  of  intersec- 
tion is  AB,  form  a  diedral 
angle,  of  which  the  line  AB 
is  the  edge,  and  the  planes  ABM,  ABN  are  the 
faces. 

392.  A  diedral  angle  may  be  acute,  right,  or  obtuse. 

If  the  two  faces  are  perpendicular  to  eacli  other,  the 
angle  is  right. 

303.     A    POLYEDRAL    AnGLE    IS 

an  angle  formed  by  the  meeting 
at  one  point  of  more  than  two 
plane  angles,  which  are  not  in 
the  same  plane. 

The  common  point  of  meeting 
of  the  planes  is  called  the  vertex, 
each  of  the  plane  angles  a  face, 
and  the  line  of  common  section  of  any  two  of  the  planes 
an  edge  of  the  polyedral  angle. 

Thus  the  three  plane  angles  ASB,  BSC,  CSA  form 
a  polyedral  angle,  whose  vertex  is  S,  whose  faces  are  the 
plane  angles,  and  whose  edges  are  the  sides,  AS,  BS,  CS, 
of  the  same  angles.- 

394.  A  polyedral  angle  formed  by  three  faces  is  called 
a  triedral  angle  ;  by  four  faces,  a  tetraedral ;  by  five  faces, 
a  pentaedral,  &q. 


Proposition  I.  —  Theorem. 

395.  A  straight  line  cannot  he  partly  in  a  plane,  and 
partly  out  of  it. 
For,  by  the  definition  of  a  plane  (Art.  10),  a  straight 


BOOK    VII. 


167 


line  wliicli  has  two  points  in  common  with  a  plane  lies 
ivholly  in  that  plane. 

396.  Scholium.  To  determine  whether  a  surface  is  a 
3lane,  apply  a  straight  line  in  different  directions  to  that 
surface,  and  ascertain  whether  the  line  throughout  its 
rvhole  extent  touches  the  surface. 


Proposition  II.  —  Theorem. 

397.  Tivo  straight  lines  which  intersect  each  other  lie 
n  the  same  plane  and  determine  its  position. 

Let  A  B,  A  C  be  two  straight  lines 
vliich  intersect  each  other  in  A ;  then 
hese  lines  will  be  in  the  same  plane. 

Conceive  a  plane  to  pass  through 
I B,  and  to  be  turned  about  A  B, 
mtil  it  pass  through  the  point  C  ; 
hen,  the  two  points  A  and  C  being  in  this  plane,  the  line 
^C  lies  wholly  in  it  (Art.  10).  Hence,  the  position  of 
he  plane  is  determined  by  the  condition  of  its  containing 
he  two  straight  lines  A B,  AC. 

398.  Cor.  1.  A  triangle,  ABC,  or  three  points.  A,  B, 
^,  not  in  a  straight  line,  determine  the  position  of  a 
►lane. 

399.  Cor.  2.  Hence,  also,  two 
parallels,  A  B,  CD,  determine 
he  position  of  a  plane ;  for, 
[rawing  the  secant  E  F,  the 
>lane  of  the  two  straight  lines 
LB,  E  F  is  that  of  the  parallels 
LB,  CD. 

Proposition  III.  —  Theorem. 

400.  If  two  planes  cut  each  other ^  their  common  section 
s  a  straight  line. 


168 


ELEMENTS   OF   GEOMETRY. 


Let  the  two  planes  A B,  CD  cut  each 
other,  and  let  E,  F  be  two  points  in  their 
common  section.  Draw  the  straight  line 
E  F.  Now,  since  the  points  E  and  F  arc  in 
the  plane  A  B,  and  also  in  the  plane  CD, 
the  straight  line  E  F,  joining  E  and  F, 
must  be  wholly  in  each  plane,  or  is  com- 
mon to  both  of  them.  Therefore,  the 
common  section  of  the  two  planes  AB, 
CD   is  a  straight  line. 


B 


D 


Proposition  TV.  —  Theorem. 

401.  If  a  straig-ht  line  is  perpendicular  to  each  of  tiao 
straight  lines,  at  their  point  of  intersection,  it  is  perpen- 
dicular to  the  plane  in  ivhich  the  tioo  lines  lie. 

Let  the  straight  line  AB  be 
perpendicular  to  each  of  the 
straight  lines  CD,  E  F,  at  B,  the 
point  of  their  intersection,  and 
MN  the  plane  in  which  the  lines 
C  D,  E  F  lie  ;  then  will  A  B  be 
perpendicular  to  the  plane  M  N. 

Through  the  point  B  draw  any  straight  line,  B  G,  in  tlie 
plane  M  N  ;  and  through  any  point  G  draw  D  G  F,  meet- 
ing the  lines  CD,  E  F  in  such  a  manner  that  D  G  shall 
be  equal  to  G F  (Frob.  XXVIII.  Bk.  V.).  Join  AD, 
A  G,  A  F. 

The  line  D  F  being  divided  into  two  equal  parts  at  the 
point  G,  the  triangle  DBF  gives  (Prop.  XIV.  Bk.  IV.) 

B  p^  +  bT)^  =  2  B^^  +  2  GFl 
The  triangle  D  AF,  in  like  manner,  gives 


A F'  +  A D-  =  2  A  Cr  +  2  G  F'. 
Subtracting  the  first  equation  from  the  second,  and  ob- 


BOOK  VII.  1G9 

serving  that  the  triangles  ABF,  ABD,  each  being  right- 
angled  at  B,  give 

AF'  — BF'  =  AB',     and     AD'  — BD^=AB^, 
we  shall  have 

AB'  +  AB'  =  2  AG'  —  2  BGl 
Therefore,  by  taking  the  halves  of  both  members,  we  have 

AB'  =  AG'— BG',      or      AG' =  AB' +  BG' ; 
hence,  the  triangle  A  B  G  is  right-angled  at  B,  and  the 
side  A  B  is  perpendicular  to  B  G. 

In  the  same  manner,  it  may  be  shown  that  AB  is  per- 
pendicular to  any  other  straight  line  in  the  plane  M  N, 
which  it  may  meet  at  B  ;  therefore  A  B  is  perpendicular 
to  the  plane  MN  (Art.  388). 

402.  Scholium.  Thus  it  is  evident,  not  only  that  a 
straight  line  may  be  perpendicular  to  all  the  straight  lines 
which  pass  tlirough  its  foot,  in  a  plane,  but  it  always  must 
be  so  whenever  it  is  perpendicular  to  two  straight  lines 
drawn  in  the  plane ;  which  shows  the  accuracy  of  the 
first  definition  (Art.  388). 

403.  Cor,  1.  The  perpendicular  AB  is  shorter  than 
any  oblique  line  A  G ;  therefore  it  measures  the  shortest 
distance  from  the  point  A  to  the  plane  M  N. 

404.  Cor.  2.  From  any  given  point,  B,  in  a  plane,  only 
one  perpendicular  to  that  plane  can  be  drawn.  For  if 
there  could  be  two,  conceive  a  plane  to  pass  through  them, 
intersecting  the  plane  MN  in  BG;  the  two  perpendiculars 
would  then  be  perpendicular  to  the  straight  line  B  G  at 
the  same  point,  and  in  the  same  plane,  which  is  impossible 
(Prop.  XIII.  Cor.,  Bk.  I.). 

It  is  also  impossible  to  let  fall  from  a  given  point  out 
of  a  plane  two  perpendiculars  to  that  plane.  For,  suppose 
A  B,  A  G  to  be  two  such  perpendiculars,  then  the  triangle 
A  B  G  will  have  two  right  angles,  A  B  G,  A  G  B,  which 
is  impossible  (Prop.  XXVIII.  Cor.  3,  Bk.  I.). 


ELEMENTS   OF   GEOMETRY. 


Proposition  Y.  —  Theorem. 

405.  Oblique  lines  drawn  from  a  point  to  a  plane  at 
equal  distances  from  a  perpendicular  dravm  from  the  same 
point  to  it,  are  equal;  and  of  tiuo  oblique  lines  unequally 
distant  from  the  perpendicular,  the  more  remote  is  the 
longer. 

Let  A  B  be  perpendicular 
to  the  plane  M  N ;  and  A  C, 
AD,  AE  be  oblique  lines, 
from  the  point  A,  meeting 
the  plane  at  equal  distances, 
B  C,  B  D,  B  E,  from  the  per- 
pendicular ;  and  A  P  a  line 
meeting  the  plane  more  remote  from  the  perpendicular ; 
then  will  A  C,  A  D,  A  E  be  equal  to  each  other,  and  AF 
be  longer  than  A  C. 

For,  the  angles  ABC,  ABD,  ABE  being  right  angles, 
and  the  distances  BC,  BD,  BE  being  equal  to  each  other, 
the  triangles  ABC,  ABD,  ABE  have  in  each  an  equal 
angle  contained  by  equal  sides  ;  consequently  they  are 
equal  (Prop.  V.  Bk.  I.)  ;  therefore,  the  hypothenuscs,  or 
the  oblique  lines  A  C,  AD,  AE,  are  equal  to  each  other. 

In  like  manner,  since  the  distance  B  F  is  greater  than 
BC,  or  its  equal  BE,  the  oblique  line  AF  must  be  greater 
than  AE,  or  its  equal  A  C  (Prop.  XIV.  Bk.  I.). 

406.  Cor.  All  the  equal  oblique  lines  AC,  AD,  AE, 
&c.  terminate  in  the  circumference  of  a  circle,  C  D  E,  de- 
scribed from  B,  the  foot  of  the  perpendicular,  as  a  centre ; 
therefore,  a  point,  A,  being  given  out  of  a  plane,  the  point 
B,  at  which  the  perpendicular  let  fall  from  it  would  meet 
that  plane,  may  be  found  by  taking  upon  the  plane  three 
points,  C,  D,  E,  equally  distant  from  the  point  A,  and 
then  finding  the  centre  of  the  circle  which  passes  through 
these  points  ;  this  centre  will  be  the  point  B  required. 


BOOK    VII. 


171 


407.  Scholium.  The  angle  A  C  B  is  called  the  inclina- 
tion of  the  oblique  line  AC  to  the  plane  M  N ;  which 
inclination  is  evidently  equal  with  respect  to  all  such 
lilies,  AC,  AD,  A E,  as  are  equally  distant  from  the  per- 
pendicular ;  for  all  the  triangles  A  C  B,  A  D  B,  A  E  B, 
&c.  are  equal  to  each  other. 


Proposition  YI.  —  Theorem. 

408.  ff  from  the  foot  of  a  perpendicular  a  straight  line 
be  draivn  at  right  angles  to  any  straight  line  of  the  plane, 
and  a  straight  line  be  draivn  from  the  point  of  intersection 
to  any  point  of  the  perpendicular,  this  last  line  will  be 
perpendicular  to  the  line  of  the  plane. 

Let  A  B  be  perpendicular  to 
the  plane  M  N,  and  B  D  a 
straight  line  drawn  through 
B,  cutting  at  right  angles  the 
straight  line  C  E  in  the  plane  ; 
draw  the  straight  line  AD  from 
the  point  of  intersection,  D,  to 
any  point.  A,  in  the  perpendicular  AB;  and  AD  will  be 
perpendicular  to   CE. 

For,  take  DE  equal  to  D  C,  and  join  BE,  BC,  AE,  AC. 
Since  D  E  is  equal  to  D  C,  the  two  right-angled  triangles 
B  D  E,  B  D  C  are  equal,  and  the  oblique  line  B  E  is  equal 
to  B  C  (Prop.  V.  Bk.  I.)  ;  and  shice  B  E  is  equal  to  B  C, 
the  oblique  line  AE  is  equal  to  A  C  (Prop.  Y.  Bk.  I.)  ; 
therefore  the  line  AD  has  two  of  its  points,  A  and  D, 
equally  distant  from  tlie  extremities  E  and  C  ;  hence, 
AD  is  a  perpendicular  to  EC,  at  its  middle  point,  D 
(Prop.  XY.  Cor.,  Bk.  I.). 

409.  Cor.  It  is  also  evident  that  C  E  is  perpendicular 
to  the  plane  of  the  triangle  xl  B  D,  since  C  E  is  perpendic- 
ular at  the  same  time  to  the  two  straight  lines  A  D  and 
BD  (Prop.  lY.). 


172 


ELEMENTS   OF   GEOMETRY. 


Proposition  VII.  —  Theorem. 

410.  If  a  straight  line  is  perpendicular  to  a  plane,  every 
plane  ivhich  passes  through  that  line  is  also  perpendicular 
to  the  plane. 

Let  A  B  be  a  straight  line 
perpendicular  to  the  plane 
M  N ;  then  will  any  plane, 
A  C,  passing  througli  A  B, 
be  perpendicular  to  MN. 

For,  let  C  D  be  the  inter- 
section of  the  planes  A  C, 
MN;  in  the  plane  MN  draw 
E  F,  through  the  point  B,  perpendicular  to  C  D  ;  then  the 
line  A  B,  being  perpendicular  to  the  plane  M  N,  is  perpen- 
dicular to  each  of  the  two  straight  lines  CD,  EF  (Art. 
388).  But  the  angle  ABE,  formed  by  the  two  perpen- 
diculars AB,  EF  to  their  common  section,  CD,  measures 
the  angle  of  the  two  planes  A  0,  MN  (Art.  891)  ;  hence, 
since  that  angle  is  a  right  angle,  the  two  planes  are  per- 
pendicular to  each  other. 

411.  Cor.  When  three  straight  lines,  as  AB,  CD,  EF, 
are  perpendicular  to  each  other,  each  of  those  lines  is  per- 
pendicular to  the  plane  of  the  other  two,  and  the  three 
planes  are  perpendicular  to  each  other. 


Proposition  Y III .  —  Theorem  . 

412.  If  two  planes  are  perpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them,  perpendicular  to  their 
common  section,  will  he  perpendicular  to  the  other  plane. 

Let  AC,  MN  be  two  planes  perpendicular  to  each 
other,  and  let  the  straight  line  AB  be  drawn  in  the  plane 
A  C  perpendicular  to  the  common  section  C  D  ;  then  will 
A  B  be  perpendicular  to  the  plane  M  N. 


BOOK    VII. 


173 


For,  in  the  plane  MN, 
draw  E  F,  through  the  point 
B,  perpendicular  to  CD ;  then, 
since  the  planes  AC,  M  N 
are  perpendicular,  the  angle 
A  B  E  is  a  right  angle  (Art. 
391)  ;  therefore  the  line  A  B 
is  perpendicular  to  the  two 
straight  lines  C  D,  E  F,  at  the  point  of  their  intersection  ; 
hence  it  is  perpendicular  to  their  plane,  MN  (Prop.  IV.). 

413.  Cor.  If  the  plane  A  C  is  perpendicular  to  the 
plane  M  N,  and  if  at  a  point  B  of  the  common  section  we 
erect  a  perpendicular  to  the  plane  M  N,  that  perpendicular 
will  be  in  the  plane  AC.  For,  if  not,  there  may  be  drawn 
in  the  plane  A  C  a  line,  A  B,  perpendicular  to  the  common 
section  C  D,  which  would  be  at  the  same  time  perpendicu- 
lar to  the  plane  M  N.  Hence,  at  the  same  point  B  there 
would  be  two  perpendiculars  to  the  plane  MN,  which 
is  impossible  (Prop.  IV.  Cor.  2). 


Proposition  IX.  —  Theorem. 

414.  If  two  planes  ivhich  cut  ea,ch  other  are  perpendic- 
ular to  a  third  plane,  their  common  section  is  perpendicu- 
lar to  the  same  plane. 

Let  the  two  planes  C  A, 
D  A,  which  cut  each  other 
in  the  straight  line  A  B,  be 
each  perpendicular  to  the 
plane  M  N  ;  then  will  their 
common  section  A  B  be  per- 
pendicular to  M  N. 

For,  at  the  point  B,  erect 


My 


a  perpendicular  to  the  plane  MN;  that  perpendicular 
must  be  at  once  in  the  plane  C  A  and  in  the  plane  D  A 
(Prop.  VIII.  Cor.)  ;  hence,  it  is  their  common  section,  A  B. 


15* 


174  ELEMENTS  OP  GEOMETRY. 

Proposition  X.  —  Theorem. 

415.  If  one  of  two  parallel  straight  lines  is  perpendic- 
ular to  a  plane,  the  other  is  also  perpendicular  to  the  same 
plane. 

Let   AB,  CD   be   two   parallel  A 

straight  lines,  of  which  A  B  is  per- 
pendicular to  the  plane  MN;  then 
will  C  D  also  be  perpendicular  to  it. 

For,  pass  a  plane    through  the 
parallels  A B,  CD,  cutting  the  plane 
MN  in  the  straight  line  BD.     In 
the  plane  M  N  draw  the  straight  line  E  F,  at  right  angles 
with  BD;  and  join  AD. 

Now,  E  F  is  perpendicular  to  the  plane  A  B  D  C  (Prop. 
VI.  Cor.)  ;  therefore  the  angle  C  D  E  is  a  right  angle  ; 
but  the  angle  C  D  B  is  also  a  right  angle,  since  A  B  is  per- 
pendicular to  B  D,  and  C  D  parallel  to  A  B  (Prop.  XXII. 
Cor.,  Bk.  I.)  ;  therefore  the  line  C  D  is  perpendicular  to 
the  two  straight  lines  E  F,  B  D  ;  hence  it  is  perpendicular 
to  their  plane,  MN  (Prop.  IV,). 

416.  Cor.  1.  Conversely,  if  the  straight  lines  AB,  CD 
are  perpendicular  to  ihe  same  plane,  M  N,  they  must  be 
parallel.  For,  if  they  be  not  so,  draw,  through  the  point 
D,  a  line  parallel  to  AB  ;  this  parallel  will  be  perpendic- 
ular to  the  plane  M  N  ;  hence,  through  the  same  point  D 
more  than  one  perpendicular  may  be  erected  to  the  same 
plane,  which  is  impossible  (Prop.  IV.  Cor.  2). 

417.  Cor.  2.  Two  lines,  A  and  B,  parallel  to  a  third, 
C,  are  parallel  to  each  other ;  for,  conceive  a  plane  per- 
pendicular to  the  line  C ;  the  lines  A  and  B,  being  parallel 
to  C,  will  be  perpendicular  to  the  same  plane  ;  hence,  by 
the  preceding  corollary,  they  will  be  parallel  to  each  other. 

The  three  lines  are  supposed  to  be  not  in  the  same 
plane  ;  otherwise  the  proposition  would  be  already  demon- 
strated (Prop.  XXIV.  Bk.  I.). 


BOOK   VII.  175 

Proposition  XI. — Theorem. 

418.  If  a  straight  line  without  a  plane  is  parallel  to  a 
line  loithin  the  plane ^  it  is  parallel  to  the  plane. 

Let  the  straight  line  A  B,  with-  A  B 

out  the  plane  M  N,  be  parallel  to 
the  line  C  D  in  that  plane  ;  then 
will  AB  be  parallel  to  the  plane 
M  N. 

Conceive  a  plane  A  B  C  D  to 
pass  through  the  parallels  AB,  CD.  Now,  if  the  line 
A  B,  wliich  lies  in  the  plane  A  B  C  D,  could  meet  the 
plane  M  N,  it  could  only  be  in  some  point  of  the  line 
CD,  the  common  section  of  the  two  planes  ;  but  the  line 
A  B  cannot  meet  C  D,  since  they  are  parallel  (Art.  17)  ; 
til  ere  fore  it  will  not  meet  the  plane  M  N  ;  hence  it  is  par- 
allel to  that  plane  (Art.  389). 

Proposition  XII.  —  Theorem. 

419.  If  tivo  planes   are    perpendicular   to   the   same 
straight  line,  the?/  are  parallel  to  each  other. 

Let  the  planes  M  N, 


M/ 

1 

I 

I 

■■-■"/ 

/n 

/ 

/ 

I 

5 

/ 

PQ,be  each  perpendic- 
ular to  the  straight  line      O.; -■  "' 
A  B  ;  then  will  they  be 
parallel  to  each  other. 

For,  if  they  can  meet, 
on  being  produced,  let 
O  be  one  of  their  com-  Q 

mon  points  ;  and  join  0  A,  0  B.  The  line  A  B,  which  is 
perpendicular  to  the  plane  M  N,  is  perpendicular  to  the 
straight  line  OA,  drawn  through  its  foot  in  that  plane 
(Art.  388).  For  the  same  reason,  AB  is  perpendicular 
to  B  0.  Therefore  O  A  and  0  B  are  two  perpendiculars 
let  fall  from  the  same  point,  0,  upon  the  same  straight 
line,   AB,   which    is    impossible    (Prop.   XIIL   Bk.  I.). 


1T6 


ELEMENTS   OP   GEOMETRY. 


Therefore,  the  planes  M  N,  P  Q  cannot  meet  on  being 
produced ;  hence  they  are  parallel  to  each  other. 

Proposition  XIII. — Theorem. 

420.  If  tioo  parallel  planes  are  cut  hy  a  third  plane ^  the 
tvjo  intersections  are  parallel. 

Let    the   two   parallel    planes  F 

MN  and  P  Q  be  cut  by  the  plane 
E  F  G  H,  and  let  their  intersec- 
tions with  it  be  E  F,  G  H  ;  then 
E  F  is  parallel  to  G  H. 

For,  if  the  lines  E  F,  G  H,  ly- 
ing in  the  same  plane,  were  not 
parallel,  they  would  meet  each 
other  on  being  produced  ;  therefore  the  planes  MN,  PQ, 
•ju  which  those  lines  are  situated,  would  also  meet,  which 
is  impossible,  since  these  planes  are  parallel. 


M/ 


N 


Proposition  XIY. — Theorem. 

421.  A  straight  line  which  is  perpendicular  to  one  of 
two  parallel  planes,  is  also  perpendicular  to  the  other  plane. 

Let  M  N,  P  Q  be  two  parallel 
planes,  and  A  B  a  straight  line 
perpendicular  to  the  plane  M  N  ; 
then  A  B  is  also  perpendicular  to 
the  plane  PQ. 

Draw  any  line,  B  C,  in  the  plane 
P  Q  ;  and  through  the  lines  A  B, 
B  C,  conceive  a  plane,  ABC,  to 
pass,  intersecting  the  plane  M  N  in  A  D  ;  the  intersection 
A  D  will  be  parallel  to  B  C  (Prop.  XIJL).  But  the  line 
A  B,  being  perpendicular  to  the  plane  M  N,  is  perpendic- 
ular to  the  straight  line  A  D  ;  consequently  it  will  be  per- 
pendicular to  its  parallel  BC  (Prop.  XXII.  Cor.,  Bk.  I.). 


BOOK  VII.  177 

Hence  the  line  A  B,  being  perpendicular  to  any  line,  B  C, 
drawn  through  its  foot  in  the  plane  P  Q,  is  consequently 
perpendicular  to  the  plane  PQ  (Art.  388). 

Proposition  X Y.  —  Theorem. 

422.  Parallel  straight  lines  included  between  tivo  par- 
allel planes  are  equal. 

Let  E  F,  G  H  be  two  parallel  

straight  planes,  included  between  /  ^, — ^"^  i       / 

two  parallel  planes,  M  N,  P  Q  ; 
then  E  F  and  G  H  are  equal. 

For,  through  the  parallels  EF, 
G  H  conceive  the  plane  E  F  G  H 
to  pass,  intersecting  the  parallel     /  '  / 

planes  in  EG,  F  H.     The  inter-  ^ 

sections  EG,  F  H  are  parallel  to  each  other  (Prop. 
XIII.)  ;  and  E  F,  G  H  are  also  parallel ;  consequently 
the  figure  E  F  G  H  is  a  parallelogram  ;  hence  E  F  is  equal 
to  GH  (Prop.  XXXI.  Bk.  I.). 

423.  Cor.  Two  parallel  planes  are  everyivhere  equi- 
distant. For,  if  E  F,  G  H  are  perpendicular  to  the  two 
planes  M  N,  P  Q,  they  will  be  parallel  to  each  other 
(Prop.  X.  Cor.  1)  ;  and  consequently  equal. 

Proposition  XVI. — Theorem. 

424.  If  two  angles  not  in  the  same  plane  have  their 
sides  parallel  and  lying  in  the  same  direction.,  these  an- 
gles will  be  equal.,  and  their  planes  ivill  be  parallel. 

Let  B  A  C,  E  D  F  be  two  tri-  ^[,- 

angles,  lying  in  different  planes, 
M  N  and  PQ,  having  their  sides 
parallel  and  lying  in  the  same 
direction ;  then  the  angles  BAG, 
EDF  will  be  equal,  and  their 
planes,  MN,  PQ,  be  parallel. 


ITS 


ILKIMKNTS    OF    ()  KOM  KTI!  V. 


M, 

/ 

TT 

TT 

/ 

i 

/ 

".if 

^ 

?c 

/. 

y 

1 

;  " 

\ 

^. 

1 

~7 

/ 

eU- 

\ 

i 

A 

For,  tako  A 15  equal  to  K  I), 
and  A  C  cuiual  to  J)  F  ;  and  join 
BC,  EF,  JiE,  AD,  OF.  Sinco 
A  B  is  equal  and  parallel  to 
E  D,  tlio  liguro  A  B  F  I)  is  a 
parallolograni  (Prop.  XXXIII. 
Bk.  1.)  ;  thoroforo  A  D  is  equal 
and  parallel  to  B  K.  For  a  similar  reason,  CF  is  equal 
and  parallel  to  A  1);  licnco,  also,  BE  is  equal  and  parallel 
to  C  F  ;  henee  the  figure  B  C  F  E  is  a  parallelogram,  and 
the  side  B  C  is  e(iual  and  ])arallel  to  E  F ;  therefore  the 
triangles  B  A  C,  E  I)  F  have  their  sides  equal,  eaeh  to 
each  ;  henee  the  angle  B  A  C  is  equal  to  the  angle  E  D  F. 

Again,  the  j)lane  BAG  is  ])arallel  to  the  plane  E  D  F. 
For,  if  not,  sup|)ose  a  i)lane  to  pass  through  the  j)oint  A, 
parallel  to  E  I)  F,  meeting  the  lines  BE,  C  F,  in  points 
diderent  from  B  and  C,  for  instanee  G  and  II.  Then  the 
three  lines  G  E,  AD,  II F  will  bo  equal  (Prop.  XV.). 
But  the  three  lines  B  E,  A  D,  C  F  are  already  known  to 
lui  ecpial ;  henee  B  E  is  equal  to  G  E,  and  J I  F  is  ecpial  to 
(/  F,  whieh  is  absurd  ;  henee  the  plane  B  A  C  is  parallel 
to  the  plane  E  D  F. 

425.  Cor,  If  two  parallel  ])lanes  M  N,  P  Q,  are  met  by 
two  other  planes,  A  B  E  D,  A  C  F  D,  the  angles  B  A  C, 
E  1)  F,  formed  by  the  interseetions  of  the  parallel  planes, 
are  equal  ;  for  the  interseetion  A 15  is  ])arallel  to  ED,  and 
AC  to  DF  (Prop.  Xlll.)  ;  therefore  the  angle  BAG  is 
equal  to  the  angle  E  1)  F. 

Proposition  XVII.  —  Theorem. 

420.  If  three  straight  lines  not  in  the  same  plane  are 
equal,  and  parallel^  the  triangles  formed  bj/  joining-  the  ex- 
trcinitirs  of  these  lines  ivill  be  equals  and  their  planes  will 
be  pa  rail  eL 

Lot  BE,  AD,  GF  bo  three  equal  and  parallel  straight 
linos,  not  in  the  same  plane,  and  let  B  A  C,  E  D  F  be  two 


BOOK    VII.  179 

triangles  formed  by  joining  the 
cxtrcniitics  of  tlicsc  lines  ;  then 
will  these  triangles  be  equal, 
and  their  planes  parallel. 

For,  since  B  E  is  equal  and 
parallel  to  AD,  the  figure 
ABED  is  a  parallelogi-am  ; 
hence,  the  side  A  B  is  equal  and  parallel  to  D  E  (Prop. 
XXXIII.  Bk.  I.).  For  a  like  reason,  the  sides  BC,  EF 
are  equal  and  parallel ;  so  also  are  AC,  D  F  ;  hence,  the 
two  triangles  B  A  C,  E  D  F,  having  their  sides  equal,  are 
themselves  equal  (Prop.  XVIII.  Bk.  I.)  ;  consequently, 
as  showu  in  the  last  proposition,  their  planes  are  parallel. 

Proposition  XVIII.  —  Theorem. 

427.  If  tioo  straight   lines  are  cnt  by  three  parallel 
planes^  they  will  be  divided  proportionally. 

Let  the  straight  line  A  B  meet 
tlie  parallel  planes,  M  N,  P  Q,  R  S, 
at  the  points  A,  E,  B  ;  and  the 
straight  line  C  D  meet  the  same 
planes  at  the  points  C,  F,  D  ;  then 
will 

AE:EB::CF:FD. 

Draw  the  line  AD,  meeting  the 
plane  P  Q  in  G,  and  draw  A  C,  E  G,  B  D.  Then  the  two 
parallel  planes  PQ,  RS,  being  cut  by  the  plane  A  B  D, 
tlie  intersections  E  G,  B  D  are  parallel  (Prop.  XIII.)  ; 
and,  in  the  triangle  A  BD,  Wo  have  (Prop.  XVII.  Bk.  IV.), 

AETeB:  :  AG:  GD. 

In  like  manner,  the  intersections  A  C,  G  F  being  paral- 
lel, in  the  triangle  ADC,  we  have 

AG:  GD  :  :  CF:FD; 


1  A/r-n 

/■m.'r 

/  V ^ 

F  / 

LvO 

-^x 

1 ''' : 

U 

180  ELEMENTS   OP   GEOMETRY. 

hence,  since  the  ratio  A  Gr  :  G  D  is  common  to  both  pro- 
portions, we  have 

AE  :  EB  :  :  OF  :  ED. 

Proposition  XIX.  —  Theorem. 

428.  TJie  sum  of  any  two  of  the  plane  angles  which 
form  a  triedral  angle  is  greater  than  the  third. 

The  proposition  requires  dem-  g 

onstration  only  when  the  plane 
angle,  which  is  compared  to  the  . 

sum  of  the  other  two,  is  greater  / 

than  either  of  them.  / 

Let  the   triedral  angle  whose        / 
vertex  is  S  be  formed  by  the  three  "•-.....         \       /'-^^ 

plane  angles  ASB,  ASC,  BSC;  """"--\X 

and  suppose  the  angle  A  S  B  to 

be  greater  than  either  of  the  other  two ;  then  the  angle 
A  S  B  is  less  than  the  sum  of  the  angles  ASC,  BSC. 

In  the  plane  ASB  make  the  angle  BSD  equal  to 
BSC;  draw  the  straight  line  A  D  B  at  pleasure  ;  make 
S  C  equal  S  D,  and  draw  A  C,  B  C. 

The  two  sides  B  S,  S  D  are  equal  to  the  two  sides  B  S, 
S  C,  and  the  angle  B  S  D  is  equal  to  the  angle  BSC; 
therefore  the  triangles  BSD,  BSC  are  equal  (Prop.  V. 
Bk.  I.)  ;  hence  the  side  B  D  is  equal  to  the  side  B  C. 
But  AB  is  less  than  the  sum  of  A  C  and  BC;  taking 
B  D  from  the  one  side,  and  from  the  other  its  equal,  B  C, 
there  remains  A D  less  than  AC.  The  two  sides  A S,  S D 
of  the  triangle  A  S  D,  are  equal  to  the  two  sides  A  S,  S  C, 
of  the  triangle  ASC,  and  the  third  side  A  D  is  less  than 
the  third  side  A  C  ;  hence  the  angle  A  S  D  is  less  than 
the  angle  ASC  (Prop.  XVII.  Bk.  I.).  Adding  B  S  D  to 
one,  and  its  equal,  BSC,  to  the  other,  we  shall  have  the 
sum  of  A  S  D,  B  S  D,  or  A  S  B,  less  than  the  sum  of  A  S  C, 
BSC.      ' 


BOOK   VII.  181 

Proposition  XX.  —  Theorem. 

429.  The  sum  of  the  plane  angles  which  form  any  poly- 
edral  angle  is  less  than  four  right  angles. 

Let  the  polyedral  angles  whose 
vertex  is  S  be  formed  by  any  number 
of  plane  angles,  A  S  B,  B  S  C,  C  S  D, 
&G. ;  the  sum  of  all  these  plane  angles 
is  less  than  four  right  angles. 

Let  the  planes  forming  the  poly- 
edral angle  be  cut  by  any  plane, 
ABCDEF.     From   any   point,  0,  B  C 

in  this  plane,  draw  the  straight  lines  A  0,  BO,  CO,  DO, 
E  0,  F  0.  The  sum  of  the  angles  of  tlie  triangles  A  SB, 
BSC,  (fee.  formed  about  the  vertex  S,  is  equal  to  the  sum 
of  the  angles  of  an  equal  number  of  triangles  AOB,  BOC, 
&c.  formed  about  the  point  0.  But  at  the  point  B  the  sum 
of  the  angles  ABO,  0  B  C,  equal  to  ABC,  is  less  than 
the  sum  of  the  angles  A  B  S,  SBC  (Prop.  XIX.)  ;  in 
the  same  manner,  at  the  point  C  we  have  the  sum  of 
B  C  0,  0  C  D  less  than  the  sum  of  B  C  S,  S  C  D  ;  and  so 
with  all  the  angles  at  the  points  D,  E,  <fec.  Hence,  the 
sum  of  all  the  angles  at  the  bases  of  the  triangles  whose 
vertex  is  0,  is  less  than  the  sum  of  all  the  angles  at  the 
bases  of  the  triangles  whose  vertex  is  S ;  therefore,  to 
make  up  the  deficiency,  the  sum  of  the  angles  formed 
about  the  point  0  is  greater  tlian-  the  sum  of  the  angles 
formed  about  the  point  S.  But  the  sum  of  the  angles 
about  the  point  0  is  equal  to  four  right  angles  (Prop.  IV. 
Cor.  2,  Bk.  L)  ;  therefore  the  sum  of  the  angles  about  S 
must  be  less  than  four  right  angles. 

430.  Scholium.  This  demonstration  supposes  that  the 
polyedral  angle  is  convex  ;  that  is,  that  no  one  of  the 
faces  would,  on  being  produced,  cut  the  polyedral  angle  ; 
if  it  were  otherwise,  the  sum  of  the  plane  angles  would 
no  longer  be  limited,  and  might  be  of  any  magnitude. 

16 


182 


ELEMENTS   OP   GEOMETRY. 


Proposition  XXI. — Theorem. 

431.  If  two  triedral  angles  are  formed  by  plane  angles 
which  are  equal  each  to  each,  the  planes  of  the  equal  an- 
gles will  be  equally  inclined  to  each  other. 

Let  the  two  triedral  an- 
gles whose  vertexes  are  S 
and  T,  have  the  angle  ASC 
equal  to  D  T  F,  the  angle 
A S B  equal  to  DTE,  and 
the  angle  B  S  C  equal  to 
ETF;  then  will  the  incli- 
nation of  the  planes  ASC,  ASB  be  equal  to  that  of  the 
planes  D  T  F,  DTE. 

For,  take  S  B  at  pleasure ;  draw  B  0  perpendicular  to 
the  plane  ASC;  from  the  point  0,  at  which  the  perpen- 
dicular meets  the  plane,  draw  OA,  0  C,  perpendicular  to 
S  A,  S  C  ;  and  join  A  B,  B  C.  Next,  take  TE  equal  S  B ; 
draw  E  P  perpendicular  to  the  plane  DTE;  from  the 
point  P  draw  PD,  PF,  perpendicular  respectively  to  TD, 
T  F  ;  and  join  D  E,  E  F. 

The  triangle  S  A  B  is  right-angled  at  A,  and  the  trian- 
gle T  D  E  at  D  ;  and  since  the  angle  A  S  B  is  equal  to 
DTE,  we  have  S  B  A  equal  to  T  E  D.  Also,  S  B  is  equal 
to  T  E  ;  therefore  the  triangle  S  AB  is  equal  to  T  D  E  ; 
hence  S  A  is  equal  to  T  D,  and  A  B  is  equal  to  D  E. 

In  like  manner  it  may  be  shown  that  S  C  is  equal  to 
T  F,  and  B  C  is  equal  to  E  F.  We  can  now  show  that  the 
quadrilateral  ASCO  is  equal  to  the  quadrilateral  DTFP; 
for,  place  the  angle  ASC  upon  its  equal  D  T  F  ;  since 
S  A  is  equal  to  T  D,  and  S  C  is  equal  to  T  F,  the  point  A 
will  fall  on  D,  and  the  point  C  on  F  ;  and,  at  the  same 
time,  AO,  which  is  perpendicular  to  S  A,  will  fall  on  D  P, 
which  is  perpendicular  to  TD,  and,  in  like  manner,  CO 
on  F  P ;  wherefore  the  point  0  will  fall  on  the  point  P, 
and  A  0  will  be  equal  to  D  P. 


BOOK  vir.  183 

But  the  triangles  A  0  B,  D  P  E  are  right-angled  at  0 
and  P;  tiie  hypotenuse  AB  is  equal  to  D  E,  and  the  side 
A  0  is  equal  to  D  P  ;  hence  the  two  triangles  are  equal 
(Prop.  XIX.  Bk.  I.)  ;  and,  consequently,  the  angle  OAB 
is  equal  to  the  angle  PDE.  The  angle  OAB  is  the  incli- 
nation of  the  two  planes  A  S  B,  A  S  C ;  and  the  angle 
PDE  is  that  of  the  two  planes  DTE,  DTP;  hence,  those 
two  inclinations  are  equal  to  each  other. 

432.  Scholium  1.  It  must,  however,  be  observed,  that 
the  angle  A  of  the  right-angled  triangle  0  A  B  is  properly 
the  inclination  of  the  two  planes  A  S  B,  A  S  C  only  when 
the  perpendicular  B  0  falls  on  the  same  side  of  S  A  with 
S  C  ;  for  if  it  fell  on  the  other  side,  the  angle  of  the  two 
planes  would  be  obtuse,  and  joined  to  the  angle  A  of  the 
triangle  0  A  B  it  would  make  two  right  angles.  But,  in 
the  same  case,  the  angle  of  the  two  planes  DTE,  D  T  F 
would  also  be  obtuse,  and  joined  to  the  angle  D  of  the 
triangle  D  P  E  it  would  make  two  right  angles  ;  and  the 
angle  A*  being  thus  always  equal  to  the  angle  D,  it 
would  follow  in  the  same  manner  that  the  inclination  of 
the  two  planes  A  S  B,  A  S  C  must  be  equal  to  that  of  the 
two  planes  DTE,  DTP. 

433.  Scholium  2.  If  two  triedral  angles  are  formed  by 
three  plane  angles  respectively  equal  to  each  other,  and 
if  at  the  same  time  the  equal  or  homologous  angles  are 
similarly  situated^  the  two  angles  are  equal.  For,  by  the 
proposition,  the  planes  which  contain  the  equal  angles  of 
tlie  triedral  angles  are  equally  inclined  to  each  other. 

434.  Scholium  3.  When  the  equal  plane  angles  forming 
tlie  two  triedral  angles  are  not  similarly  situated^  these 
angles  are  equal  in  all  their  constituent  parts,  but,  not 
admitting  of  superposition,  are  said  to  be  equal  by  sym- 
metry,  and  are  called  symmetrical  angles. 


BOOK  VIII 


POLYEDRONS. 


DEFINITIONS. 


435.  A  PoLYEDRON  is  a  solid,  or  volume,  bounded  by 
planes. 

The  bounding  planes  are  called  the  face$  of  the  polye- 
dron  ;  and  the  lines  of  intersection  of  the  faces  are  called 
tlie  edges  of  the  polyedron. 

436.  A  Prism  is  a  polyedron  having 
two  of  its  faces  equal  and  parallel  pol- 
ygons, and  the  other  faces  parallelo- 
grams. 

The  equal  and  parallel  polygons  are 
called  the  bases  of  the  prism,  and  the 
parallelograms  its  lateral  faces.  The 
lateral  faces  taken  together  constitute 
the  lateral  or  convex  surface  of  the 
prism. 

Thus  the  polyedron  ABCDE-K  is  a  prism,  having 
for  its  bases  the  equal  "and  parallel  polygons  ABODE, 
F  G  H  I  K,  and  for  its  lateral  faces  the  parallelograms 
A  B  G  F,  B  C  H  G,  <fec. 

The  principal  edges  of  a  prism  are  those  which  join  the 
corresponding  angles  of  the  bases  ;  as  A  F,  B  G,  <fec. 

437.  The  altitude  of  a  prism  is  a  perpendicular  drawn 
from  any  point  in  one  base  to  the  plane  of  the  other. 

438.  A  Right  Prism  is  one  whose  principal  edges  are 
perpendicular  to  the  planes  of  its  bases.      Each  of  the 


BOOK   VIII. 


185 


edges  is  then  equal  to  the  altitude  of  the  prism.  Every 
other  prism  is  oblique^  and  has  each  edge  greater  than  the 
altitude. 

439.  A  prism  is  triangular^  quadrangular,  pentangular, 
hexangular,  &c.,  according  as  its  base  is  a  triangle,  a 
quadrilateral,  a  pentagon,  a  hexagon,  &c. 


440.  A  Parallelopipedon  is  a  prism 
whose  bases  are  parallelograms  ;  as  the 
prism  A  B  C  D  -  H. 

The  parallelopipedon  is  rectangular 
when  all  its  faces  are  rectangles  ;  as  the 
parallelopipedon  A  B  C  D  -  H. 


441.  A  Cube,  or  Regular  Hexaedron, 
is  a  rectangular  parallelopipedon  having 
all  its  faces  equal  squares  ;  as  the  paral-     A 
lelopipedon  A  B  C  D  -  H. 


442.  A  Pyramid  is  a  polyedron  of 
which  one  of  the  faces  is  any  polygon, 
and  all  the  others  are  triangles  meeting 
at  a  common  point. 

The  polygon  is  called  the  base  of  the 
pyramid,  the  triangles  its  lateral  faces, 
and  the  point  at  which  the  triangles  meet 
its  vertex.     The  lateral   faces  taken  to-  B 

gether   constitute   the  lateral  or   convex  surface  of  the 
pyramid. 

Thus  the  polyedron  ABCDE-Sisa  pyramid,  having 
for  its  base  tlie  polygon  A  B  C  D  E,  for  its  lateral  faces  tlie 
triangles  ASB,  BSC,  CSD,  &c.,  and  for  its  vertex  the 
pohit  S. 

16* 


186  ELEMENTS   OF   GEOMETRY. 

443.  The  Altitude  of  a  pyramid  is  a  perpendicular 
drawn  from  the  vertex  to  the  plane  of  the  base. 

444.  A  pyramid  is  triang-ular,  quadrangular,  &g.,  ac- 
cording as  its  base  is  a  triangle,  a  quadrilateral,  &c. 

445.  A  Right  Pyramid  is  one  whose  base  is  a  regular 
polygon,  and  the  perpendicular  drawn  from  the  vertex  to 
the  base  passes  through  the  centre  of  the  base,  in  this 
case  tlie  perpendicular  is  called  the  axis  of  the  pyramid. 

446.  The  Slant  Height  -of  a  right  pyramid  is  a  line 
drawn  from  the  vertex  to  the  middle  of  one  of  the  sides 

of  the  base. 

'i 

447.  A  Frustum  of  a  pyramid  is  the  part  of  the  pyra- 
mid included  between  the  base  and  a  plane  cutting  the 
pyramid  parallel  to  the  base. 

448.  The  Altitude  of  the  frustum  of  a  pyramid  is  the 
perpendicular  distance  between  its  parallel  bases. 

449.  The  Slant  Height  of  a  frustum  of  a  right  pyra- 
mid is  that  part  of  the  slant  height  of  the  pyramid  which 
is  intercepted  between  the  bases  of  the  frustum. 

450.  The  Axis  of  the  frustum  of  a  pyramid  is  that  part 
of  the  axis  of  the  pyramid  which  is  intercepted  between 
the  bases  of  the  frustum. 

451.  The  Diagonal  of  a  polyedron  is  a  line  joining  the 
vertices  of  any  two  of  its  angles  which  are  not  in  tlie  same 
face. 

452.  Similar  Polyedrons  are  those  which  are  bounded 
by  the  same  number  of  similar  faces,  and  have  their  poly- 
edral  angles  respectively  equal. 

453.  A  Regular  Polyedron  is  one  whose  faces  are  all 
equal  and  regular  polygons,  and  whoso  polyedral  angles 
are  all  equal  to  each  otlicr. 


B       C 


BOOK    YIII.  187 


Proposition  I. — Theorem. 

454.  The  convex  surface  of  a  rig-ht  prism  is  equal  to 
the  perimeter  of  its  base  multiplied  by  its  altitude. 

Let  ABCDE-K  be  a  right  prism;  K 

then  will  its  convex  surface  be  equal         p 
to  the  perimeter  of  its  base, 

AB+.BC  +  CD  +  DE  +  EA, 

multiplied  by  its  altitude  A  F. 

For,  the  convex  surface  of  the  prism 
is  equal  to  the  sum  of  the  parallelo- 
grams AG,  BH,  CI,  DK,  EF  (Art. 
436).  Now,  the  area  of  each  of  those  parallelograms  is 
equal  to  its  base,  AB,  B  C,  CD,  &c.,  multiplied  by  its 
altitude,  AF,  B  G,  CH,  &c.  (Prop.  Y.  Bk.  IV.).  But 
the  altitudes  AF,  BG,  C  H,  (fcc.  are  each  equal  to  A F, 
the  altitude  of  the  prism.  Hence,  the  area  of  these  paral- 
lelograms, or  the  convex  surface  of  the  prism,  is  equal  to 

(AB  +  BC  +  CD  +  DE  +  EA)X  AF; 
or  the  product  of  the  perimeter  of  the  prism  by  its  alti- 
tude. 

455.  Cor.  If  two  right  prisms  have  the  same  altitude, 
their  convex  surfaces  are  to  each  other  as  the  perimeters 
of  their  bases. 

Proposition  II.  —  Theorem. 

456.  In  everij  prism,  the  sections  formed  by  parallel 
planes  are  equal  polygons. 

Let  the  prism  ABCDE-K  be  intersected  by  the 
parallel  planes  N  P,  S  Y  ;  then  are  the  sections  NOPQR, 
STYXY  equal  polygons. 

For  the  sides  ST,  NO  are  parallel,  being  the  intersec- 
tions of  two  parallel  planes  with  a  third  plane  A  B  G  F 


188 


ELEMENTS    OF   GEOMETRY. 


(Prop.  XIII.  Bk.  YII.)  ;  these  same 
sides  ST,  NO,  are  included  between 
the  parallels  N  S,  OT,  which  are  sides 
of  the  prism  ;  hence  N  0  is  equal  to 
S  T.  For  like  reasons,  the  sides  0  P, 
P  Q,  Q  R,  &c.  of  the  section  N  0  P  Q  R, 
are  respectively  equal  to  the  sides 
TV,  YX,  XY,  &c.  of  the  section 
S  T  Y  X  Y  ;  and  since  the  equal  sides 
are  at  the  same  time  parallel,  it  fol- 
lows that  the  angles  NOP,  OP  Q,  &c. 
of  the  first  section  are  respectively 
equal  to  the  angles  STY,  TYX  of  the  second  (Prop. 
XYI.  Bk.  YII.).  Hence,  the  two  sections  NOPQR, 
S  T  Y  X  Y,  are  equal  polygons. 

457.   Cor,   Every  section  made  in  a  prism  parallel  to 
its  base,  is  equal  to  that  base. 


Proposition  III.  —  Theorem. 

458.  Two  prisms  are  equal,  when  the  three  faces  which 
form  a  triedral  angle  in  the  one  are  equal  to  those  vjhich 
form  a  triedral  angle  in  the  other,  each  to  each,  and  are 
similarly  situated. 

Let  the  two  prisms 
A.BCDE-K  and 
LMOPQ-Y  have 
the  faces  which  form 
the  triedral  angle  B 
equal  to  the  faces 
which  form  the  tri- 
edral angle  M ;  that 
is,the base  ABODE 
equal  to  the  base  L  M  N  0  P  Q,  the  parallelogram  A  B  G  F 
equal  to  the  parallelogram  LMSR,  and  the  parallelogram 
B  0  H  G  equal  to  MOTS;  then  the  two  prisms  are  equal. 


BOOK    VIII. 


189 


For,  apply  the  base 
ABODE  to  the 
equal  base  LMOPQ ; 
then,  the  triedral  an- 
gles B  and  M,  being 
equal,  will  coincide, 
since  the  plane  an- 
gles which  form  these 
triedral    angles    are  B       C  MO 

equal  each  to  each,  and  similarly  situated  (Prop.  XXI. 
Sch.  2,  Bk.  yil.)  ;  hence  the  edge  B  G  will  fall  on  its 
equal  M  S,  and  the  face  B  H  will  coincide  with  its  equal 
MT,  and  the  face  BF  with  its  equal  MR.  But  the  upper 
bases  are  equal  to  their  corresponding  lower  bases  (Art. 
436)  ;  therefore  the  bases  F  G  H I K,  R  S  T  V  Y  are  equal ; 
hence  they  coincide  with  each  other.  Therefore  H I  coin- 
cides with  TV,  IK  with  VY,  and  KF  with  YR;  and 
consequently  the  lateral  faces  coincide.  Hence  the  two 
prisms  coincide  throughout,  and  are  equal. 

459.  Cor,  Two  right  prisms,  which  have  equal  bases 
and  equal  altitudes,  are  equal. 

For,  since  the  side  AB  is  equal  to  LM,  and  the  altitude 
B  G  to  M  S,  the  rectangle  A  B  G  F  is  equal  to  the  rectan- 
gle L  M  S  R  ;  so,  also,  the  rectangle  B  G  H  C  is  equal  to 
M  S  T  0 ;  and  thus  the  three  faces  which  form  the  triedral 
angle  B,  are  equal  to  the  three  faces  which  form  the  trie- 
dral angle  M.     Hence  the  two  prisms  are  equal.^^^X-»X.^ 

Proposition  I Y.  —  Theorem.       V\C>   ^^>v 

460.  In  every  parallelopipedon  the  opposite  ./^^^Os,^-. 
equal  and  parallel,  ^^^>^  tj 

Let  ABCD-H  be  a  parallelopipedon ;  then  its  opp5^ 
site  faces  are  equal  and  parallel. 

The  bases  ABC D,  EFGH  are  equal  and  parallel  (Art. 
436),  and  it  remains  only  to  be  shown  that  the  same  is 


190 


ELEMENTS   OF   GEOMETRY. 


H 


X> 


B 


true  of  any  two  opposite  lateral  faces,  as 
B  CGF,  ADHE.  Now,  since  the  base 
A  B  C  D   is   a   parallelogram,   the   side 

A  D  is  equal  and  parallel  to  B  C.     For  r) ^G 

a  similar  reason,  AE  is  equal  and  par- 
allel to  B  F ;  hence  the  angle  DAE  is 
equal  to  the  angle  C  B  F  (Prop.  XYI. 
Bk.  VII.),  and  the  planes  DAE,  CBF 
are  parallel ;  hence,  also,  the  parallelogram  B  C  G  1^^  is 
equal  to  the  parallelogram  A  D  H  E.  In  the  same  way, 
it  may  be-  shown  that  the  opposite  faces  A  B  F  E,  D  C  G  II 
are  equal  and  parallel. 

461.  Cor.  Any  two  opposite  faces  of  a  parallelopipe- 
don  may  be  assumed  as  its  bases,  since  any  face  and  the 
cue  opposite  to  it  are  gqual  and  parallel. 


Proposition  V.  —  Theorem. 


G 


462.  The  diagonals  of  every  parallelopipedon  bisect 
each  other. 

Let  A  B  C  D  -  H  be  a  parallelo- 
pipedon ;  then  its  diagonals,  as 
B  H,  D  F,  will  bisect  each  other. 

For,  since  B  F  is  equal  and  par- 
allel to  D  H,  the  figure  B  F  H  D  is 
a  parallelogram  ;  hence  the  diago- 
nals B  H,  D  F  bisect  each  other  at 
the  point  0  (Prop.  XXXIV.  Bk.  I.).  In  the  same  man- 
ner it  may  be  shown  that  the  two  diagonals  A  G  and  C  E 
bisect  each  other  at  the  point  0 ;  hence  the  several  diag- 
onals bisect  each  other. 

463.  Scholium.  The  point  at  which  the  diagonals  mu- 
tually bisect  each  other  may  be  regarded  as  the  centre  of 
the  parallelopipedon. 


E 

H 

\ 

.--•■•'■■■;/K 

/    ";".;.::::.,iT) 

\ 

7      ...-• -■  r 

B 


BOOK    VIII. 


191 


Proposition  VI.  —  Theorem. 

464.  Amj  paraUelopipedon  may  be  divided  into  tico 
equivalent  triangular  prisms  by  a  plane  passing  through 
its  opposite  diagonal  edges. 

Let  any  paraUelopipedon,  ABC  D-H,  q 

be  divided  into  two  prisms,  A  B  C  -  G, 
AD  C-G,  by  a  plane,  A  C  G  E,  passing 
through  opposite  diagonal  edges;  then 
will  the  two  prisms  be  equivalent. 

Through  the  vertices  A  and  E,  draw 
the  planes  A  K  L  M,  E  N  0  P,  perpen- 
dicular to  the  edge  AE,  and  meeting 
BE,  CG,  DH,  the  three  other  edges 
of  the  paraUelopipedon,  in  the  points 
K,  L,  M,  and  in  N,  0,  P.  The  sections  A  K  L  M,  E  N  O  P 
are  equal,  since  they  are  formed  by  planes  perpendicular 
to  the  same  straight  lines,  and  hence  parallel  (Prop.  II.). 
They  are  parallelograms,  since  the  two  opposite  sides  of 
the  same  section,  AK,  LM,  are  the  intersections  of  two 
parallel  planes,  A  B  F  E,  D  C  G  H,  by  the  same  plane, 
AKLM  (Prop.  XIII.  Bk.  VII.). 

For  a  like  reason,  the  figure  A  M  P  E  is  a  parallelo- 
gram ;  so,  also,  are  A  K  N  E,  K  L  0  N,  L  M  P  0,  the  other 
lateral  faces  of  the  solid  AKLM-P;  consequently,  this 
solid  is  a  prism  (Art.  436)  ;  and  this  prism  is  right, 
since  the  edge  AE  is  perpendicular  to  the  plane  of  its 
base.  This  right  prism  is  divided  by  the  plane  ALOE 
into  the  two  right  prisms  AKL-0,  AML-0,  which, 
liaving  equal  bases,  A  K  L,  A  M  L,  and  the  same  altitude, 
AE,  are  equal  (Prop.  III.  Cor.). 

Now,  since  AEHD,  AEPM  are  parallelograms,  the 
sides  D  H,  MP,  being  each  equal  to  AE,  are  equal  to  each 
other ;  and  taking  away  the  common  part,  D  P,  there  re- 
mains D  M  equal  to  H  P.  In  the  same  manner  it  may 
be  shown  that  C  L  is  equal  to  G  0. 


192 


ELEMENTS   OP   GEOMETRY. 


Conceive  now  E  P  0,  the  base  of  the 
solid  E  P  0  -  G,  to  be  applied  to  its 
equal  A  M  L,  the  point  P  falling  upon 
M,  and  the  point  0  upon  L  ;  the  edges 
G  0,  H  P  will  coincide  with  their  equals 
C L,  DM,  since  they  are  all  perpen- 
dicular to  the  same  plane,  A  K  L  M. 
Hence  tlie  two  solids  coincide  through- 
out, and  are  therefore  equal.  To  each 
of  these  equals  add  the  solid  ADC-P, 
and  the  right  prism  AML-0  is  equivalent  to  the  prism 
ADC-G. 

In  the  same  manner,  it  may  be  proved  that  the  riglit 
prism  AKL-0  is  equivalent  to  the  prism  ABC-G.  Tlie 
two  riglit  prisms  A  K  L  -  0,  A  M  L  -  0  being  equal,  it  fol- 
lows that  two  triangular  prisms,  ABC-G,  ADC-G, 
are  equivalent  to  each  other. 

465.  Cor.  Every  triangular  prism  is  half  of  a  parallelo- 
pipedon  having  the  same  triedral  angle,  with  the  same 
edges. 

Proposition  VII.  —  Theorem. 


466.  Two  parallelopipedons,  having  a  common  lower 
base,  and  their  upper  bases  in  the  same  plane  and  between 
the  same  parallels,  are  equivalent  to  each  other. 

Let  the  two  parallelo- 
pipedons A  G,  A  L  have 
tlie  commonbase  ABCD, 
and  their  upper  bases, 
EFGH,  IKLM,  in 
the  same  plane,  and  be- 
tween the  same  paral- 
lels, EK,  HL;  then  the 
parallelopipedons  will  be  equivalent. 


BOOK   VIII.  193 

There  may  be  throe  cases,  according  as  E I  is  greater  or 
less  than,  or  equal  to,  EF;  but  the  demonstration  is  the 
same  for  each. 

Since  AE  is  parallel  to  BF,  and  HE  to  GF,  the  pla]io 
anjrle  A  E  I  is  equal  to  B  F  K,  H  E  I  to  G  F  K,  and  H  E  A 
to  G  F  B.  Of  these  six  plane  angles,  the  three  first  form 
the  polyedral  angle  E,  the  three  last  the  polyedral  angle 
F ;  consequently,  since  these  plane  angles  are  equal  each 
to  each,  and  similarly  situated,  the  polyedral  angles,  E,  F, 
must  be  equal.  Now  conceive  the  prism  A  E  I  -  M  to  be 
applied  to  the  prism  B  F  K  -  L  ;  the  base  A  E  I,  being 
placed  upon  the  base  B  F  K,  will  coincide  witli  it,  since 
they  are  equal ;  and,  since  the  polyedral  angh^  E  is  equal 
to  the  polyedral  angle  F,  the  side  E  II  will  fall  upon  its 
equal,  F  G.  But  the  base  A  E  I  and  its  edge  I']  II  deter- 
mine the  prism  A  E  I-M,  as  the  base  B  F  K  and  its  edge 
FG  determine  tlje  prism  BFK-L  (Prop.  III.)  ;  hcuce 
the  two  prisms  coincide  throughout,  and  therefore  are 
equal  to  each  other. 

Take  away,  now, from  the  whole  solid  AELC,  the  prism 
AEI-M,  and  tliere  will  remain  the  parallelopipcdon  AL; 
and  take  away  from  the  same  solid  A  L  the  prism  BFK-L, 
and  there  will  remain  the  parallelopipcdon  A  G ;  hence 
the  two  parallelopipedons  A  L,  A  G  are  equivalent. 

Proposition  YIII.  —  Theorem. 

467.  Tvjo  parallelopipedons  having-  the  same  base  and 
the  same  altitude  are  equivalent. 

Let  the  two  parallelopipedons  AG,  A L  have  tlie  com- 
mon base  ABCD,  and  the  same  altitude  ;  then  will  the 
two  parallelopipedons  be  equivalent. 

For,  the  upper  bases  EFGH,  IKLM  being  in  the  same 
plane,  produce  the  edges  E  F,  HG,  L  K,  IM,  till  ]>y  their 
intersections  they  form  the  parallelogram  N  0  P  Q  ;  this 
parallelogram  is  eq\ial  to  eitlier  of  the  bases  I  L,  E  G,  and 

17 


194 


ELEMENTS   OF   GEOMETRY. 


is  between  the  same  par- 
allels ;  hence  N  0  P  Q  is 
equal  to  the  common 
base  A  B  C  D,  and  is 
parallel  to  it. 

Now,  if  a  third  paral- 
lelopipedon  be  conceived, 
which,  with  tlie  same 
lower  base  A  B  C  D,  has 
for  its  upper  base  NOPQ, 

this  tliird  parallelopipe-  A"^ ^B 

don  will  be  equivalent  to  the  parallelopipedon  A  G,  since 
the  lower  base  is  the  same,  and  tlie  upper  bases  lie  in 
the  same  plane  and  between  the  same  parallels,  GQ,  FN 
(Prop.  VII.). 

For  the  same  reason,  this  third  parallelopipedon  will 
also  be  equivalent  to  the  parallelopipedon  A  L  ;  hence  the 
two  parallelopipedons  AG,  AL,  which  have  the  same  base 
and  the  same  altitude,  are  equivalent. 


Proposition  IX.  —  Theorem. 

468.  Any  oblique  'parallelopipedon  is  equivalent  to  a 
rectangular  parallelopipedon  hauing-  the  same  altitude 
and  an  equivalent  base. 

Let  A  G  be  any  paral-  H  G 

lelopipedon  ;  then  A  G 
will  be  equivalent  to  a 
rectangular  parallelopip- 
edon having  the  same 
altitude  and  an  equiv- 
alent base. 

From  the  points  A, 
B,  C,  D,draw  AI,  BK, 
C  L,  D  M,  perpendicular 
to  the  lower  base,  and 
equal  in  altitude  to  A  G ;  there  will  thus  be  formed  the 


BOOK    VIII. 


195 


parallelopipedon  AL,  equivalent  to  A  G  (Prop.  YIII.), 
and  liaving  its  lateral  faces,  AK,  B  L,  <fec.,  rectangular. 
Now,  if  the  base  ABCD  is  a  rectangle,  AL  will  be  a  rec- 
tangular parallelopipedon  equivalent  to  A  G. 

But  if  ABCD  is  not  a  rectangle,       j^jq  ^p 

draw  A  0,  B  N,  each  perpendicular  to 
CD;  also  0  Q,  N  P,  each  perpendicular 
to  the  base  ;  then  we  shall  have  a  rec- 
tangular parallelopipedon  A  B  N  0  -  Q. 
For,  by  construction,  the  bases  ABNO, 
I  K  P  Q  are  rectangles  ;  so,  also,  are 
the  lateral  faces,  the  edges  A  I,  0  Q, 
&c.  being  perpendicular  to  the  plane 
of  the  base  ;  therefore  the  solid  A  P  is 
a  rectangular  parallelopipedon.  But  the  two  parallelopip- 
edons  A  P,  A  L  may  be  considered  as  having  the  same 
base,  A  BKI,  and  tlie  same  altitude,  AO  ;  hence  they  are 
equivalent.  Hence  the  parallelopipedon  A  G,  which  was 
shown  to  be  equivalent  to  the  parallelopipedon  AL,  is  also 
equivalent  to  tlie  rectangular  parallelopipedon  A  P,  having 
the  same  altitude,  A  I,  and  a  base,  ABNO,  equivalent  to 
the  base  ABCD. 


Proposition  X. — Theorem. 

469.    Two   rectangular  parallelopipedons,  tvhich 
the  same  base,  are  to  each  other  as  their  altitudes. 

Let  the  two  parallelopipedons  AG,       -^ 
AL  have  the  same  base,  ABCD  ;  then 
they  are  to  each  other  as  their  altitudes, 
AE,  AL 

First.  Suppose  tlie  altitudes  AE,  AI 
are  to  each  other  as  two  whole  numbers ; 
for  example,  as  15  is  to  8.  Divide  A  E 
into  15  equal  parts,  of  which  A I  will 
contain  8.  Through  .^•,  y,  z,  <fec.,  the 
points  of  division,  conceive   planes   to 


have 


196 


ELEMENTS   OP   GEOMETRY. 


pass  parallel  to  the  common  base. 
These  planes  will  divide  the  solid  A  G 
into  15  small  parallelopipedons,all  equal 
to  each  other,  having  equal  bases  and 
equal  altitudes ;  equal  bases,  since  every 
section,  as  1  K  L  M,  parallel  to  the  base 
A  B  C  D,  is  equal  to  that  base  (Prop. 
II.),  and  equal  altitudes,  since  the  alti- 
tudes are  the  equal  divisions  Ax,  xy^ 
y  2r,  &c.  But  of  those  15  equal  parallel- 
opipedons,  8  are  contained  in  A  L  ;  hence  the  parallelo- 
pipedon  A  G  is  to  the  parallelopipedon  A  L  as  15  is  to  8, 
or,  in  general,  as  the  altitude  A  E  is  to  the  altitude  A  I. 

Secondly,  If  the  ratio  of  A  E  to  A I  cannot  be  exactly 
expressed  by  numbers,  we  shall  still  have  tlie  proportion, 

Solid  A  G  :  Solid  A  L  :  :  A  E  :  A  I. 
For,  if  this  proportion  is  not  correct,  suppose  we  have 

Solid  A  G  :  Solid  A  L  :  :  A  E  :  A  0  greater  than  A  I. 

Divide  A  E  into  equal  parts,  each  of  which  shall  be  less 
than  I  0  ;  there  will  be  at  least  one  point  of  division,  m, 
between  I  and  0.  Let  P  represent  the  parallelopipedon, 
whose  base  is  A  B  C  D,  and  altitude  Am  ;  since  the  alti- 
tudes AE,  Am  are  to  each  other  as  two  whole  numbers, 
we  shall  have 

Solid  AG  :  P  :  :  AE  :  Am. 

But,  by  hypothesis,  we  have 

Solid  A  G  :  Solid  A  L  :  :  A  E  :  A  0  ; 
hence  (Prop.  X.  Cor.  2,  Bk.  II.), 

Solid  AL  :  P  :  :  AG  :  Am. 
But  AO  is  greater  than  Am ;  lience,  if  the  proportion  is 
correct,  the  parallelopipedon  A  L  must  be  greater  than  P. 
On  the  contrary,  however,  it  is  less  ;  consequently  the 
solid  A  G  cannot  be  to  the  solid  A  L  as  the  line  A  E  is  to 
a  line  greater  than  A  I. 


BOOK    VIII. 


197 


By  the  same  mode  of  reasoning,  it  may  be  shown  that 
tlie  fourth  term  of  the  proportion  cannot  be  less  tlian  A I ; 
therefore  it  must  be  equal  to  A  I.  Hence  rectangular 
parallelopipedons,  having  the  same  base,  are  to  each  other 
as  their  altitudes. 


Proposition  XI.  —  Theorem. 

470.    Tivo   rectangular  parallelopipedons^   having'   the 
same  altitude,  are  to  each  other  as  their  bases. 

Let   the   two  rectan- 


T 


E 


H 


\K 


i\. 


k. 


Ml 


A\ 


\ riQ 

i 

j 

I)j  I 


gular  parallelopipedons 
AG,  AK  have  the  same 
altitude,  A  E ;  then  they 
are  to  each  other  as  their 
bases. 

Place  the  two  solids 
so  that  their  faces,  BE, 
0  E,  may  have  the  com- 
mon angle  B  A  E  ;  pro- 
duce the  plane  0  N  K  L 
till  it  meets  the  plane 
D  C  G  H  in  P  Q  ;  we 
shall  thus  have  a  third 
parallelopipedon,  A  Q,  wliich  may  be  compared  with  each 
of  the  parallelopipedons  A  G,  A  K.  The  two  solids,  A  G, 
A  Q,  having  the  same  base,  A  E  H  D,  are  to  each  other  as 
their  altitudes  A  B,  A  0  (Prop.  X.)  ;  in  like  manner,  the 
two  solids  A  Q,  AK,  having  the  same  base,  A  0  L  E,  arc 
to  each  other  as  their  altitudes  AD,  AM.  Hence  we 
have  the  two  proportions, 

Solid  A  G  :  Solid  A  Q  :  :  A  B  :  A  0, 
Solid  A  Q  :  Solid  A  K  :  :  A  D  :  A  M. 
Multiplying  together  the  corresponding  terms  of  these 


B 


198^  * 


ELEMENTS    OP   GEOMETRY. 


proportions,  and  omitting,  in  the  result,  the  common  factor 
Solid  A  Q,  we  shall  have, 

Solid  A  G  :  Solid  AK::ABxAD:AOxAM. 

But  A  B  X  A  D  measures  the  base  A  B  C  D  (Prop.  lY. 
Sell.,  Bk.  lY.)  j  a^id  AO  X  AM  measures  the  base 
A  M  N  0  ;  hence  two  rectangular  parallelopipcdons  of  the 
same  altitude  are  to  each  other  as  their  bases. 


M 


Proposition  XII.  —  Theorem. 

471.  Any  tivo  rectang-ular  parallelopipedons  are  to  each 
other  as  the  product  of  their  bases  by  their  altitudes. 

Let  A  G,  A  Z  be  two       i  v  vi 

rectangular  parallelo- 
pipedons ;  then  they  are 
to  each  other  as  the 
product  of  their  bases, 
ABCD,  AMNO,  by 
their  altitudes,  A E,  AX. 

Place  the  two  solids 
so  that  their  faces,  B  E, 
0  X,  may  have  the  com- 
mon angle  B  A  E  ;  pro- 
duce the  planes  neces- 
sary for  completing  the 
third  parallelopipedon, 
A  K,  having  the  same  altitude  with  the  parallelopipedon 
A  G.     By  the  last  proposition,  we  shall  have 

Solid  A  G  :  Solid  AK::ABCD:AMNO. 

But  the  two  parallelopipedons  A  K,  A  Z,  having  the  same 
base,  A  M  N  0,  are  to  each  other  as  their  altitudes,  A  E, 
A  X  (Prop.  X.)  ;  hence  we  have 

Solid  AK  :  Solid  A  Z  :  :  A  E  :  A  X. 
Multiplying  together  the  corresponding  terms  of  these 


V      ft, 

'      ^. 

i     \ 

xi 

i  i 

^                          i  1 

Di  i 

\z      K 

> 

N       \ 

\i 

N.                   OV 

\       ■        \ 

c 


BOOK  VIII.  199 

proportions,  and  omitting,  in  the  result,  the  common  fac- 
tor Solid  A  K,  we  shall  have 

SolidKGc'.Solid^Z::A.BOJ)X^^:  AMNOx  AX. 
Hence,  any  two  rectangular  parallelopipedons  are  to  each 
other  as  the  products  of  their  bases  by  their  altitudes. 

472.  Scholium  1.  We  are  consequently  authorized  to 
assume,  as  the  measure  of  a  rectajigular  parallelopipedon, 
the  product  of  its  base  by  its  altitude  ;  in  other  words,  the 
product  of  its  three  dimensions.  But  by  the  product  of 
two  or  more  lines  is  always  meant  the  product  of  the  num- 
bers which  represent  them  ;  those  numbers  themselves 
being  determined  by  the  particular  linear  unit,  which  may 
be  assumed  as  the  standard.  It  is  necessary,  therefore,  in 
comparing  magnitudes,  that  the  measuring  unit  be  the 
same  for  each  of  the  magnitudes  compared. 

473.  Scholium  2.  The  measured  magnitude  of  a  solid, 
or  volume,  is  called  its  volume,  solidit?/,  or  solid  contents. 
We  assume  as  the  unit  of  volume,  or  solidity,  the  cube, 
each  of  whose  edges  is  the  linear  unit,  and  each  of  whose 
faces  is  the  unit  of  surface. 

Proposition  XIII.  —  Theorem. 

474.  The  solid  contents  of  a  parallelopipedon,  and  of 
any  other  prism,  are  equal  to  the  product  of  its  base  by 
its  altitude. 

First.  Any  parallelopipedon  is  equivalent  to  a  rectan- 
gular parallelopipedon  having  the  same  altitude  and  an 
equivalent  base  (Prop.  IX.).  But  the  solid  contents  of  a 
rectangular  parallelopipedon  are  equal  to  the  product  of 
its  base  by  its  altitude;  therefore  the  solid  contents  of  any 
parallelopipedon  are  equal  to  the  product  of  its  base  by  its 
altitude. 

Second.  Any  triangular  prism  is  half  of  a  parallelopip- 
edon, so  constructed  as  to  have  the  same  altitude,  and  a 


200  ELEMENTS  OF  GEOMETRY. 

base  twice  as  great  (Prop.  YI.).  But  the  solid  contents 
of  the  parallelopipedon  are  equal  to  the  product  of  its  base 
by  its  altitude  ;  hence,  that  of  the  triangular  prism  is  also 
equal  to  the  product  of  its  base,  or  half  that  of  the  paral- 
lelopipedon, by  its  altitude. 

Third.  Any  prism  may  be  divided  into  as  many  trian- 
gular prisms  of  the  same  altitude,  as  there  are  triangles  in 
the  polygon  taken  for  a  base.  But  the  solid  contents  of 
each  triangular  prism  are  equal  to  the  product  of  its  base 
by  its  altitude;  and,  since  the  altitude  is  the  same  in  each, 
it  follows  that  the  sum  of  all  these  prisms  is  equal  to  the 
sum  of  all  the  triangles  taken  as  bases  multiplied  by  the 
common  altitude. 

Hence  the  solid  contents  of  any  prism  are  equal  to  the 
product  of  its  base  by  its  altitude. 

475.  Cor.  When  any  two  prisms  have  the  same  altitude, 
the  products  of  the  bases  by  the  altitudes  will  be  as  the 
bases  (Prop.  IX.  Bk.  11.)  ;  hence,  prisms  of  the  same 
altitude  are  to  each  other  as  their  bases.  For  a  like 
reason,  prisms  of  the  same  base  are  to  each  other  as  their 
altitudes. 

Proposition  XI Y.  —  Theorem. 

476.  Similar  prisms  are  to  each  other  as  the  cubes  of 
their  homologous  edges. 

Let  ABC-E,  FHI-M  e 
be  two  similar  prisms ; 
these  prisms  are  to  each 
other  as  the  cubes  of  their 
homologous  edges,  A  B 
and  FII. 

For,  from  I)  and  K,  ho- 
mologous  angles    of   the 
two  prisms,  draw  the  perpendiculars  DN,  KO,  to  the  bases 
ABC,  FHI.     Take  AK'  equal  to  F  K,  and  join  AN. 


BOOK   VIIT. 


201 


Draw  K' 0'  perpendicular  to  A N  in  the  plane  AND,  and 
K'  0  will  be  perpendicular  to  tlie  plane  ABC,  and  equal 
to  K  0,  the  altitude  of  the  prism  F  H  I  -  M.  For,  con- 
ceive the  triedral  angles  A  and  F  to  be  applied  the  one  to 
the  other  ;  the  planes  containing  them,  and  therefore  the 
perpendiculars  K'  0',  K  0,  will  coincide. 

Now,  since  the  bases  A  B  C,  F  H  I  are  similar,  we  have 
(Prop.  XXIX.  Bk.  IV.), 

Base  ABC:  Base  F  H  I  :  :  AB^  :  FH^ ; 

and,  because  of  the  similar  triangles  DAN,  K F 0,  and  of 
the  similar  parallelograms  D  B,  K  H,  we  have 

DN:KO::DA:KF::AB:FH. 

Hence,  multiplying  together  the  corresponding  terms  of 
these  proportions,  we  have 

Base  ABC  XT)^  :  Base  FRlXKO:  AB^  :  FW. 

But  the  product  of  the  base  by  the  altitude  is  equal  to  the 
solidity  of  a  prism  (Prop.  XIII.)  ;  hence 

PmmABC-E:  Pmm  F  H  I-M  :  :  AB' :  FH'. 


Proposition  XV.  —  Theorem. 

477.  The  convex  surface  of  a  rig-lU  pyramid  is 
to  the  perimeter  of  its  base,  multiplied  by  half  the 
height. 

Let  A  B  C  D  E  -  S  be  a  right  pyra- 
mid, and  S  M  its  slant  heiglit ;  then  the 
convex  surface  is  equal  to  the  perimeter 
AB  +  BC  +  CD  +  DE  +  EA  mul- 
tiplied by  J-  S  M. 

The  triangles  SAB,  SBC,  S  C  D, 
<fcc.  are  all  equal ;  for  the  sides  A  B, 
BC,  CD,  &c.  are  equal  (Art.  445),  and 
the  sides  S  A,  S  B,  S  C,  &c.,  being  ob- 
lique lines  meeting  the  base  at  equal 


equal 
slant 


202 


ELEMENTS    OF   GEOMETRY. 


distances  from  a  perpendicular  let  fall 
from  the  vertex  S  to  the  centre  of  the 
base,  are  also  equal  (Prop.  V.  Bk. 
VII.).  Hence,  these  triangles  are  all 
equal  (Prop.  XVIII.  Bk.  I.)  ;  and  the 
altitude  of  each  is  equal  to  the  slant 
height  S  M.  But  the  area  of  a  triangle 
is  equal  to  the  product  of  its  base  mul- 
tiplied by  half  its  altitude  (Prop.  VI. 
Bk.  IV.).  Hence,  the  areas  of  the  tri- 
angles SAB,  SBC,  SCD,  &c.  are  equal  to  the  sum  of  the 
bases  A  B,  B  C,  CD,  &c.  multiplied  by  half  the  common 
altitude,  S  M  ;  that  is,  the  convex  surface  of  the  pyramid 
is  equal  to  the  perimeter  of  the  base  multiplied  by  half  the 
slant  height. 

478.  Cor.  The  lateral  faces  of  a  right  pyramid  are 
equal  isosceles  triangles,  having  for  their  bases  the  sides 
of  the  base  of  the  pyramid. 


Proposition  XVI. — Theorem. 

479.  If  a  pyramid  be  cut  bi/  a  plane  parallel  to  its  base, — 

1st.    The  edges  and  the  altitude  will  be  divided  propor- 
tionally. 

2d.    The  section  ivill  be  a  polygon  similar  to  the  base. 

Let  the  pyramid  A  B  C  D  E  -  S,  whose  S 

altitude  is  SO,  be  cut  by  a  plane, 
G  H  I K  L,  parallel  to  its  base  ;  then 
will  the  edges  S  A,  S  B,  S  C,  &c.,  with 
the  altitude  SO,  be  divided  proportion- 
ally ;  and  the  section  G  H  I  K  L  will 
be  similar  to  the  base  ABODE. 

First.  Since  the  planes  ABC,  GHI 
are  parallel,  their  intersections  A  B, 
GH,  by  the  third  plane  SAB,  are 
parallel  (Prop.  XIII.  Bk.  VII.);  hence 


BOOK  vm.  20.S 

tlie  triangles  SAB,  SGH  are  similar  (Prop.  XXY.  Bk. 
IV.),  and  we  have 

SA:SG::SB:SH. 

For  the  same  reason,  we  have 

SB:  SH:  :  SC:  SI; 

and  so  on.  Hence  all  the  edges,  S  A,  S  B,  S  C,  &c.,  are 
cut  proportionally  in  G,  H,  I,  &c.  The  altitude  S  0  is 
likewise  cut  in  the  same  proportion,  at  the  point  P ;  for 
B  0  and  H  P  are  parallel ;  therefore  we  have 

SO:SP::SB:SH. 

Secondly,  Since  G  H  is  parallel  to  AB,  H  I  to  B  C,  I K 
to  C  D,  (fee.  the  angle  G  H  1  is  equal  to  A  B  C,  the  angle 
H  I K  to  B  C  D,  and  so  on  (Prop.  XVI.  Bk.  VII.).  Also, 
by  reason  of  the  similar  triangles  SAB,  S  G  H,  we  have 

AB:GH::SB:SH; 

and  by  reason  of  the  similar  triangles  SBC,  SHI,  we 
have 

SB:  SH:  :  BC  :  HI; 

hence,  on  account  of  the  common  ratio  S  B  :  S  H, 

A  B  :  G  H  :  :  B  C  :  H  I. 
For  a  like  reason,  we  have 

B  C  :  H  I  :  :  C  D  :  I K, 

and  so  on.  Hence  the  polygons  ABCDE,  GHIKL 
have  their  angles  equal,  each  to  each,  and  their  homolo- 
gous sides  proportional ;  hence  they  are  similar. 

480.  Cor.  1.  If  two  pyramids  have  the  same  altitude, 
and  their  bases  in  the  same  plane,  their  sections  made  by 
a  plane  parallel  to  the  plane  of  their  bases  are  to  each 
other  as  their  bases. 

Let  ABCPE-S,  MNO-S  be  two  pyramids,  having 
the  same  altitude,  and  their  bases  in  the  same  plane  ;  and 
let  G  H  I  K  L,  P  Q  R  be  sections  made  by  a  plane  parallel 


204 


ELEMENTS  OF  GEOMETRY. 


:  S  A  :  S  G. 


:S^^ 


to  the  plane  of  their  bases ; 
then  these  sections  are  to 
each  other  as  the  bases 
ABODE,  M  N  0. 

For,  the  two  polygons 
ABODE,  GHIKL  be- 
ing similar,  their  surfaces 
are  as  the  squares  of  the 
homologous  sides  AB,  GH 
(Prop.  XXXI.  Bk.  IV.). 
But 

AB:  GH 
Hence, 

ABODE:  GHIKL:  :SA' 

For  the  same  reason, 

M  N  O  :  P  Q  R  :  :  S  M^  :  SP^- 
But  since  GHIKL  and  P  Q  R.  are  in  the  same  plane, 
we  have  also  (Prop.  XVIII.  Bk.  VII.), 

SA:  SG::SM:  SP; 
hence 

'  ABODE:  GHIKL:  :MNO:PQR; 

therefore  the  sections  GHIKL,  PQR  are  to  each  other 
as  the  bases  ABODE,  M  N  0. 

481.  Cor.  2.  If  the  bases  A  B  0  D  E,  M  N  O  are  equiv- 
alent, any  sections,  GHIKL,  PQR,  made  at  equal  dis- 
tances from  those  bases,  are  likewise  equivalent. 

Proposition   XVII.  —  Theorem. 

482.  The  convex  surface  of  a  frustum  of  a  right  pi/ra- 
viid  is  equal  to  half  the  sum  of  the  perimeters  of  its  two 
bases,  multiplied  by  its  slant  height. 

Let  ABODE-L  be  the  frustum  of  a  riglit  pyramid, 
and  MN  its  slant  height;  then  the  convex  surface  is  equal 
to  the  sum  of  the  perimeters  of  the  two  bases  ABODE, 
G  H  I K  L,  multiplied  by  half  of  M  N. 


BOOK   VIII. 


205 


For  the  upper  base  G  H I  K  L  is  similar 
to  the  base  ABODE  (Prop.  XVI.),  and 
ABODE  is  a  regular  polygon  (Art. 
445) ;  hence  the  sides  G  H,  HI,  IK, 
K  L,  and  L  G  are  all  equal  to  each  other. 
The  angles  GAB,  A  B  H,  HBO,  &c. 
are  equal  (Prop.  XV.  Oor.),  and  the 
edges  AG,  B  H,  01,  &c.  are  also  equal 
(Prop.  XVI.)  ;  therefore  the  faces  A  H, 
B  I,  OK,  &c.  are  all  equal  trapezoids  (Art.  28),  havuig 
a  common  altitude,  M  N,  the  slant  height  of  the  frustum. 
But  the  area  of  either  trapezoid,  as  AH,  is  equal  to 
^(AB+GH)XMN  (Prop.  VII.  Bk.  IV.)  ;  hence 
the  areas  of  all  the  trapezoids,  or  the  convex  surface  of 
frustum,  are  equal  to  half  the  sum  of  the  perimeters  of  the 
two  bases  multiplied  by  the  slant  height. 


Proposition  XVIII.  —  Theorem. 

483.  Triangular  pyramids^  having  equivalent  bases  and 
the  same  altitude,  are  equivalent. 
T  S  __ 


B  B' 

Let  A  B  0  -  S,  A'  B'  0'  -  S'  be  two  triangular  pyramids, 
having  equivalent  bases,  ABO,  A'  B'  0',  situated  in  the 
same  plane  ;  and  let  them  have  the  same  altitude,  A  T  ; 
then  these  pyramids  are  equivalent. 

For,   if  the    two   pyramids    are    not    equivalent,   let 
A'  B'  0  -  S'  be  the  smaller,  and  suppose  A  X  to  be  the 

18 


206 


ELEMENTS    OF   GEOMETRY. 


altitude  of  a  prism,  which,  having  ABC  for  its  base,  is 
equal  to  thoir  difference. 

T  S 


B  B' 

Divide  the  altitude  A  T  into  equal  parts,  each  less  than 
A  X  ;  through  each  point  of  division  pass  a  plane  parallel 
to  the  plane  of  the  base,  thus  forming  corresponding 
sections  in  the  two  pyramids,  equivalent  each  to  each, 
namely,  DEF  to  D'E'F,  GHI  to  G'HT,  &c. 

Upon  the  triangles  ABC,  DEF,  GHI,  <fec.,  taken  as 
bases,  construct  exterior  prisms,  having  for  edges  the  parts 
AD,  D  G,  GK,  &c.  of  the  edge  S  A ;  in  like  manner,  on 
the  bases  D'  E'  F',  G'  H'  T,  &c.  in  the  second  pyramid, 
construct  interior  prisms,  having  for  edges  the  correspond- 
ing parts  of  S'  A'.  It  is  plain  that  the  sum  of  all  the  ex- 
terior prisms  of  the  pyramid.  A  B  C-S  is  greater  than  this 
pyramid  ;  and  also  that  the  sum  of  all  the  interior  prisms 
of  the  pyramid  A'B'C'-S'  is  less  than  this  pyramid. 
Hence,  the  difference  between  the  sum  of  all  tlie  exterior 
prisms  and  the  sum  of  all  the  interior  ones,  must  be 
greater  than  the  difference  between  the  two  pyramids 
themselves. 

Now,  beginning  with  the  bases  ABC,  A'B'C',  the 
second  exterior  prism,  D  E  F  -  G,  is  equivalent  to  the  first 
interior  prism,  D'E'F'-A',  since  they  have  equal  alti- 
tudes, and  their  bases,  DEF,  D'E'F',  are  equivalent.  For 
a  like  reason,  the  third  exterior  prism,  GHI-K,  and  tlio 
second  interior  prism,  G'HT-D',  are  equivalent;  and  so 


BOOK   VIII. 


207 


on  to  the  last  in  each  series.  Hence,  all  the  exterior 
prisms  of  the  pyramid  A  B  C  -  S,  excepting  the  first  prism, 
ABC-D,  have  equivalent  correspondhig  ones  in  the  in- 
terior prisms  of  the  pyramid  A  B  C'-S'.  Therefore  the 
prism  ABC-D  is  the  difference  hetween  the  sum  of  all 
the  exterior  prisms  of  the  pyramid  ABC-S,  and  the  sum 
of  the  interior  prisms  of  the  pyramid  A'  B'  C  -  S'.  But 
the  difference  between  these  two  sets  of  prisms  has  been 
proved  to  be  greater  than  that  of  the  two  pyramids,  which 
latter  difference  we  supposed  to  be  equal  to  tlie  prism 
A  B  C  -  X.  Hence,  the  prism  ABC-D  must  be  greater 
than  tlie  prism  A  B  C  -X,  which  is  impossible,  since  they 
have  the  same  base,  ABC,  and  the  altitude  of  the  first  is 
less  than  A  X,  the  altitude  of  the  second.  Hence,  the 
supposed  inequality  between  the  two  pyramids  cannot 
exist ;  therefore  the  two  pyramids  ABC-S,  A'  B'  C-  S', 
having  the  same  altitude  and  equivalent  bases,  are  them- 
selves equivalent. 


Proposition  XIX. — Theorem. 

484.  Ever?/  triangular  pyramid  is  a  third  part  of  a  tri- 
angular prism  having  the  same  base  and  the  same  altitude. 

Let  A  B  C  -  F  be  a  triangu- 
lar pyramid,  and  A  B  C  -  D  E  F 
a  triangular  prism  of  the  same 
base  and  the  same  altitude ; 
tlien  the  pyramid  is  one  third 
of  the  prism. 

C  ut  off  the  pyramid  A  B  C  -  F 
from  the  prism,  by  the  plane 
F  A  C  ;  there  will  remain  the 
solid  A  C  D  E  -  F,  which  may 
be  considered  as  a  quadrangu- 
lar pyramid,  whose  vertex  is  F, 
and  whose  base  is  the  parallelogram  A  C  D  E.     Draw  the 


208 


ELEMENTS   OP   GEOMETRY. 


diagonal  CE,  and  pass  the  plane 
F  C  E,  which  will  cut  the  quad- 
rangular pyramid  into  two  tri- 
angular ones,  ACE-F,  EDC-F. 
These  two  triangular  pyramids 
have  for  their  common  altitude 
the  perpendicular  let  fall  from 
F  on  the  plane  A  C  D  E  ;  they 
have  equal  bases,  since  the  tri- 
angles A  C  E,  C  D  E  are  halves  of 
the  same  parallelogram  ;  hence 
the  two  pyramids  ACE-F, 
C  D  E-F  are  equivalent  (Prop.  XYIII.).  But  the  pyra- 
mid CDE-F  and  the  pyramid  ABC-F  have  equal 
bases,  ABC,  D  E  F  ;  they  have  also  the  same  altitude, 
namely,  the  distance  between  the  parallel  planes  ABC, 
D  E  F  ;  hence  the  tw^o  pyramids  are  equivalent.  Now, 
the  pyramid  CDE-F  has  been  proved  equivalent  to 
ACE-F;  hence  the  three  pyramids  ABC-F,  CDE-F, 
A  C  E-F,  which  compose  the  whole  prism  A  B  C -D  E  F, 
are  all  equivalent;  therefore,  either  pyramid,  as  ABC-F, 
is  the  third  part  of  the  prism,  which  has  the  same  base 
and  the  same  altitude. 

485.  Cor,  1.  Every  triangular  prism  may  be  divided 
into  three  equivalent  triangular  pyramids. 

486.  Cor.  2.  The  solidity  of  a  triangular  pyramid  is 
equal  to  a  third  part  of  the  product  of  its  base  by  its 
altitude. 


Proposition  XX.  —  Theorem. 

487.  The  solidity  of  every  pyramid  is  equal  to  the  pro- 
duct of  its  base  hy  one  third  of  its  altitude. 

Let  ABCDE-S  be  any  pyramid,  whose  base  is 
ABCDE,  and  altitude  SO;  then  its  solidity  is  equal 
toABCDEx^SO. 


BOOK  VIII.  209 

Draw  the  diagonals  AC,  AD,  and 
pass  the  pliuies  SAC,  SAD  tlirough 
these  diagonals  and  the  A'ertex  S  ;  the 
polygonal  pyramid  A  B  C  D  E  -  S  will 
be  divided  into  several  triangular  pyra- 
mids, all  having  the  same  altitude,  S  0. 
But  each  of  these  pyramids  is  measured 
by  the  product  of  its  base,  B  AC,  CAD, 
D  A  E,  by  a  third  part  of  its  altitude, 
S  0  (Prop.  XIX.  Cor.  2)  ;  hence,  the 
sum  of  these  triangular  pyramids,  or  the  polygonal  pyra- 
mid A  B  C  D  E  -  S,  will  be  measured  by  the  sum  of  the 
triangles  B  A  C,  CAD,  D  A  E,  or  the  polygon  A  B  C  D  E, 
multiplied  by  one  third  of  S  0 ;  hence,  every  ])yramid  is 
measured  by  the  product  of  its  base  by  one  third  of  its 
altitude. 

488.  Cor.  1.  Every  pyramid  is  the  third  part  of  the 
prism  which  has  the  same  base  and  the  same  altitude. 

489.  Cor.  2.  Pyramids  having  the  same  altitude  are  to 
each  other  as  their  bases. 

490.  Cor.  3.  Pyramids  having  the  same  base,  or  equiv- 
alent bases,  are  to  each  other  as  their  altitudes. 

491.  Cor.  4.  Pyramids  are  to  each  other  as  the  pro- 
ducts of  their  bases  by  their  altitudes. 

492.  Scholium.  The  solidity  of  any  polyedron  may  be 
found  by  dividing  it  into  pyramids,  by  passing  planes 
through  its  vertices. 

Proposition  XXI. — Theorem. 

493.  A  fruslnm  of  a  pyramid  is  equivalent  to  the  sum 
of  three  pyramids^  having  for  their  common  altitude  the 
altitude  of  the  frustum ,  and  whose  bases  are  the  two  bases 
of  the  frustum  and  a  mean  proportional  bettveen  them. 

First.  Let  A  B  C-D  E  F  be  the  frustum  of  a  pyramid, 
whose  base  is  a  triangle.     Pass  a  plane  through  the  points 


210 


ELEMENTS   OP   GEOMETRY. 


A,  E,  C ;  it  cuts  off  the  triangular 
pyramid  A  I>  C  -  E,  whose  altitude 
is  that  of  the  frustum,'  and  whose 
base,  ABC,  is  the  lower  base  of 
the  frustum.  Pass  another  plane 
through  the  points  D,  E,  C;  it  cuts 
off  the  triangular  pyramid  D  E  F  -  C, 
whose  altitude  is  that  of  the  frus- 
tum, and  whose  base,  DEE,  is  the 
upper  base  of  the  frustum. 

There  now  remains  of  the  frus- 
tum the  pyramid  A  C  D  -  E.  Draw  E  G  parallel  to  AD  ; 
join  C  G  and  D  G.  Then,  since  E  G  is  parallel  to  AD,  it 
is  parallel  to  the  plane  A  C  D  (Prop.  XL  Bk.  VII.)  ;  and 
the  pyramid  A  C  D  -  E  is  equivalent  to  the  pyramid 
AC.  D-G,  since  they  have  the  same  base,  A  C  D,  and 
their  vertices,  E  and  G,  lie  in  the  same  straight  line  par- 
allel to  the  common  base.  But  the  pyramid  A  C  D  -  G  is 
the  same  as  the  pyramid  AGC-D,  whose  altitude  is  that 
of  the  frustum,  and  whose  base,  A  G  C,  as  will  be  proved, 
is  a  mean  proportional  between  the  bases  ABC  and  DEE. 

The  two  triangles  A  G  C,  DEE  have  the  angles  A  and 
D  equal  to  each  other  (Prop.  XVI.  Bk.  VII.)  ;  hence  we 
have  (Prop.  XXVIII.  Bk.  IV.), 

AGC:DEF::AGXAC:DEXDF; 

but  since  A  G  is  equal  to  D  E, 

AGC  :  DEE  :  :  AC  :  D  F. 

We  have,  also  (Prop.  VI.  Cor.,  Bk.  IV.), 

A  B  C  :  A  G  C  :  :  A  B  :  A  G  or  D  E. 

But  the  similar  triangles  ABC,  DEE  give 

A  B  :  D  E  :  :  A  C  :  D  F  ; 

hence  (Prop.  X.  Bk.  II.), 

ABC:AGC::AGC:DEF; 


BOOK   VIII. 


211 


tliat  is,  the  base  A  G  C  is  a  mean  proportional  between  the 
bases  A  B  C,  D  E  F  of  the  frustum. 

Secondly.  Let  G  H  I K  L  -  M  N  0  P  Q  be  the  frustum 
of  a  pyramid,  whose  base  is  any  polygon. 

Let  ABC-S  be 
a  triangular  pyramid 
liaving  the  same  alti- 
tude, and  an  equiva- 
lent base,  with  any 
polygonal  pyramid, 
G  H  I K  L  -  T  ;  these 
pyramids  are  equiva- 
lent (Prop.  XX.  Cor.  3.) 

The  bases  of  the  two  pyramids  may  be  regarded  as  situ- 
ated in  the  same  plane,  in  which  case  the  plane  MNOPQ 
produced  will  form  in  the  triangular  pyramid  a  section, 
D  E  F,  at  the  same  distance  above  the  common  plane  of 
the  bases  ;  and  therefore  the  section  D  E  F  will  be  to 
the  section  MNOPQ  as  the  base  ABC  is  to  the  base 
G  H I K  L  (Prop.  XVI.  Cor.  1)  ;  and  since  the  bases  are 
equivalent,  the  sections  will  be  so  likewise.  Hence,  the 
pyramids  MNOPQ- T,  DEF-S,  having  the  same  al- 
titude and  equivalent  bases,  are  equivalent.  For  the 
same  reason,  the  entire  pyramids  GHIKL-T,  ABC-S 
are  equivalent;  consequently,  the  frustums  GHIKL- 
M  N  0  P  Q,  A  B  C  -  D  E  F,  are  equivalent.  But  the  frustum 
A  B  C  -  D  E  F  has  been  shown  to  be  equivalent  to  the 
sum  of  three  pyramids  having  for  their  common  altitude 
the  altitude  of  the  frustum,  and  whose  bases  are  the  two 
bases  of  the  frustum,  and  a  mean  proportional  between 
tlicm.  Hence  the  proposition  is  true  of  the  frustum  of 
any  pyramid. 


Proposition  XXII. — Theorem. 

494.   Similar  pyramids  are  to  each  other  as  the  cubes 
of  their  homolog-ous  edg-es. 


212 


ELEMENTS  OP    GEOMETRY. 


Let  ABC-S  and 
DEF-S  be  two  sim- 
ilar pyramids  ;  these 
pyramids  are  to  each 
otlier  as  the  cubes  of 
their  homologous  edges 
AB  and  D  E,  or  BC 
and  EF,  &c. 

For,  the  two  pyra- 
mids being  similar,  the 
homologous  polyedral  angles  at  the  vertices  are  equal 
(Art.  452)  ;  hence  the  smaller  pyramid  may  be  so  applied 
to  the  larger,  that  the  polyedral  angle  S  shall  be  common 
to  both. 

In  that  case,  the  bases  ABC,  D  E F  will  he  parallel ; 
for,  since  the  homologous  faces  are  similar,  the  angle 
S  D  E  is  equal  to  S  A  B,  and  S  E  F  to  SBC;  hence  the 
plane  A  B  C  is  parallel  to  the  plane  D  E  F  (Prop.  XVI. 
Bk.  VII.).  Then  let  S  0  be  drawn  from  the  vertex  S 
perpendicular  to  the  plane  ABC,  and  let  P  be  the  point 
where  this  perpendicular  meets  the  plane  D  E  F.  From 
what  has  already  been  shown  (Prop.  XVI.),  we  shall  have 

SO:SP::SA:SD::AB:DE; 
and  consequently, 

^SO:  ^SP:  :  AB:I)E. 
But  the  bases  ABC,  DEF  are  similar;  hence  (Prop. 
XXIX.  Bk.  IV.), 

A  B  C  :  D  E  F  :  :  AB^  :  We\ 
Multiplying  together  the  corresponding  terms  of  these  two 
proportions,  we  have 

ABCXi^SO:DEFXiSP::  AB'  :  DEI 
Now,  A  B  C  X  1  S  O  represents  the  solidity  of  the  pyra- 
mid ABC-S,  and  D  E  F  X  ^  S  P  that  of  the  pyramid 
DEF-S  (Prop.  XX.)  ;  hence  two  similar  pyramids  are 
to  each  other  as  the  cubes  of  their  homologous  edges. 


BOOK   VIII.  213 

Proposition   XXIII.  —  Theorem. 

495.  There  can  be  no  more  than  Jive  regular  pohjedrons. 

For,  since  regular  polyedroiis  have  equal  regular  poly- 
gons for  tlieir  faces,  and  all  their  polyedral  angles  equal, 
there  can  be  but  few  regular  pulyedrons. 

Firsl.  If  tlie  faces  are  equilateral  triangles,  polyedrons 
may  be  formed  of  them,  having  each  polyedral  angle  con- 
tained by  tliree  of  these  triangles,  forming  a  solid  bounded 
by  fjur  equal  equilateral  triangles  ;  or  hy  four,  forming  a 
solid  bounded  by  eight  equal  equilateral  triangles  ;  or  by 
Jire,  forming  a  solid  bounded  by  twenty  equal  equilateral 
triangles.  No  others  can  be  formed  with  equilateral 
triangles.  For  six  of  these  angles  are  equal  to  four  riglit 
angles,  and  cannot  form  a  polyedral  angle  (Prop.  XX. 
Bk.  VII.). 

Secondly/,  If  the  faces  are  squares,  their  angles  may  be 
arranged  by  threes,  forming  a  solid  bounded  by  six  equal 
squares.  Four  angles  of  a  square  are  equal  to  four  right 
angles,  and  cannot  form  a  polyedral  angle. 

Thirdly/.  If  the  faces  are  regular  pentagons,  their  angles 
may  be  arranged  by  threes,  forming  a  solid  bounded  by 
twelve  equal  and  regular  pentagons. 

We  can  proceed  no  farther.  Three  angles  of  a  regular 
hexagon  are  equal  to  four  right  angles  ;  three  of  a  hepta- 
gon are  greater.  Hence,  there  can  be  formed  no  more 
than  five  regular  polyedrons,  —  three  with  equilateral  tri- 
angles, one  with  squares,  and  one  with  pentagons. 

496.  Scholium.  The  regular  polyedron  bounded  by  four 
equilateral  triangles  is  called  a  tetraedron  ;  the  one 
bounded  by  eight  is  called  an  octaedron  ;  tlie  one  bound- 
ed by  twenty  is  called  an  icosaedron.  The  regular  polj^e- 
dron  bounded  by  six  equal  squares  is  called  a  hexaedron, 
or  CUBE  ;  and  tlie  one  boundod  by  twelve  equal  and  regu- 
lar pentagons  is  called  a  dodecaedron. 


BOOK    IX. 


THE    SPHERE,  AND   ITS   PROPERTffiS. 


DEFINITIONS. 

497.  A  Sphere  is  a  solid,  or  Yolume,  bounded  by  a 
curved  surface,  all  points  of  which  are  equally  distant 
from  a  point  within,  called  the  centre. 

The  sphere  may  be  con- 
ceived to  be  formed  by  the 
revolution  of  a  semicircle, 
DAE,  about  its  diameter, 
D  E,  which  remains  fixed. 

498.  The  Radius  of  a  sphere 
is  a  straiglit  line  drawn  from 
the  centre  to  any  point  in 
surface,  as  the  line  C  B. 

The  Diameter,  or  Axis,  of  a  sphere  is  a  line  passing 
through  the  centre,  and  terminated  both  ways  by  the  sur- 
face, as  the  line  D  E. 

Hence,  all  the  radii  of  a  sphere  are  equal ;  and  all  the 
diameters  are  equal,  and  each  is  double  the  radius. 

499.  A  Circle,  it  will  be  shown,  is  a  section  of  a  sphere. 
A  Great  Circle  of  the  sphere  is  a  section  made  by  a 

plane  passing  through  the  centre,  and  having  the  centre 
of  the  sphere  for  its  centre  ;  as  the  section  A  B,  whose 
centre  is  C. 

500.  A  Small  Circle  of  the  sphere  is  any  section  made 
by  a  plane  not  passing  througli  tlie  centre. 

501.  The  Pole  of  a  circle  of  tlie  sphere  is  a  point  in  the 


BOOK    IX. 


215 


surface  equally  distant  from  every  point  in  the  circumfer- 
ence of  the  circle. 

502.  It  will  be  shown  (Prop.  V.)  that  every  circle, 
great  or  small,  has  two  poles. 

503.  A  Plane  is  tangent  to  a  sphere,  when  it  meets  the 
sphere  in  but  one  point,  however  far  it  may  be  produced. 

504.  A  Spherical  Angle 
is  the  difference  in  the  direc- 
tion of  two  arcs  of  great  cir- 
cles of  the  sphere  ;  as  A  E  D, 
formed  by  the  arcs  E  A,  D  E. 

It  is  the  same  as  the  angle 
resulting  from  passing  two 
planes  through  those  arcs  ;  as 
the  angle  formed  on  the  edge 
EP,  by  the  planes  ExVF,  EDP. 

505.  A  Spherical  Triangle  is  a  portion  of  the  surface 
of  a  sphere  bounded  by  three  arcs  of  great  circles,  each 
arc  being  less  than  a  semi-circumference  ;  as  A  E  D. 

These  arcs  are  named  the  sides  of  the  triangle  ;  and  the 
angles  which  their  planes  form  with  each  other  arc  the 
ang-les  of  the  triangle. 

506.  A  spherical  triangle  takes  the  name  of  ri^ht-ang-Ied, 
isosceles^  equilateral,  in  the  same  cases  as  a  plane  triangle. 

507.  A  Spherical  Polygon  is  a  portion  of  the  surface 
of  a  sphere  bounded  by  several  arcs  of  great  circles. 

508.  A  LuNE  is  a  portion  of 
the  surface  of  a  sphere  com- 
prehended between  semi-cir- 
cumferences of  two  great  cir- 
cles;  as  AIGBDF. 

509.  A  Spherical  Wedge, 
or  Ungula,  is  that  portion  of  a 
sphere  comprehended  between 


2i(J 


ELEMENTS   OF   GEOMETiiY. 


two  great  semicircles  having  a 
comnioii  diameter. 

510.  A  Zone  is  a  portion  of 
the  surface  of  a  sphere  cut  ofT 
by  a  plane,  or  comprehended 
between  two  parallel  planes  ; 
as  EIFK-A,  or  CGDII- 
EIFK. 

511.  A  Spherical  Segment 
is  a  portion  of  the  sphere  cut  off  by  a  plane,  or  compre- 
hended between  two  parallel  ))lanes. 

612.  The  Altitude  of  a  Zone  or  of  a  Spherical  Seg- 
ment is  the  perpendicular  distance  between  the  two  parallel 
planes  which  comprehend  the  zone  or  segment. 

In  case  the  zone  or  segment  is  a  portion  of  the  sphere 
cut  off,  one  of  the  planes  is  a  tangent  to  tlie  sphere. 

513.  A  Spherical  Sector  is  a  solid  described  by  the 
revolution  of  a  circular  sector,  in  the  same  manner  as  the 
semicircle  of  which  it  is  a  part,  by  revolving  round  its 
diameter,  describes  a  sphere. 

514.  A  Spherical  Pyramid  is  a  portion  of  the  sphere 
comprehended  between  the  planes  of  a  polyedral  angle 
whose  vertex  is  the  centre. 

The  base  of  the  pyramid  is  the  spherical  polygon  inter- 
cepted by  the  same  planes. 


Proposition  I.  —  Theorem. 

515.  Every  section  of  a  sphere  made  by  a  plane  is  a 
circle. 

Let  ABE  be  a  section  made  by  a  plane  in  the  sphere 
whose  centre  is  C.  From  the  centre,  C,  draw  CD  per- 
pendicular to  tlie  plane  ABE;  and  draw  the  lines  C  A, 
C  B,  C  E,  to  different  points  of  the  curve  ABE,  which 
bounds  the  section. 


BOOK   IX. 


217 


The  oblique  lines  C  A,  C  B,  C  E 
are  equal,  being  radii  of  the  sphere ; 
therefore  they  are  equally  distant 
from  the  perpendieular,  CD  (Prop. 
y.  Cor.,  Bk.  VII.).  Hence,  the 
lines  DA,  D  B,  D  E,  and,  in  like 
manner,  all  the  lines  drawn  from 
I)  to  the  boundary  of  the  section, 
are  ccpial ;  and  therefore  the  section  ABE  is  a  circle 
whose  centre  is  D. 

516.  Cor.  1.  If  the  section  passes  through  the  centre 
of  the  sphere,  its  radius  will  be  the  radius  of  the  sphere  ; 
hence  all  great  circles  are  cijual. 

517.  Chr.  2.  Two  great  circles  always  bisect  each  other. 
For,  since  the  two  circles  have  the  same  centre,  tlicir  com- 
mon intersection,  passing  through  the  centre,  must  \)Q  a 
common  diameter  bisecting  both  circles. 

518.  Cor.  3.  Every  great  circle  divides  the  splierc  and 
its  surface  into  two  equal  parts.  For  if  the  two  hemi- 
splieres  were  separated,  and  afterwards  placed  on  the  com- 
mon base,  with  their  convexities  turned  the  same  way,  the 
two  surfaces  would  exactly  coincide. 

519.  Cor.  4.  The  centre  of  a  small  circle,  and  that  of 
the  sphere,  are  in  a  straight  line  perpendicular  to  the 
j)lane  of  the  small  circle. 

520.  Cor.  5.  ^  Small  circles  are  less  according  to  their 
distance  from  the  centre  ;  for,  the  greater  the  distance 
CD,  the  smaller  the  chord  AB,  the  diameter  of  the  small 
circle  ABE. 

521.  Cor.  6.  The  arc  of  a  great  circle  may  be  made  to 
pass  through  any  two  points  on  the  surface  of  a  sphere ; 
for  the  two  given  points  and  the  centre  of  the  sj)here  deter- 
mine the  position  of  a  plane.  If,  however,  the  two  given 
points  be  the  extremities  of  a  diameter,  these  two  points 

19 


218 


ELEMENTS   OF   GEOMETRY, 


and  the  centre  would  be  in  a  straight  line,  and  any  num- 
ber of  great  circles  may  be  made  to  pass  through  the  two 
given  points. 

Proposition  II.  —  Theorem. 

522.  Any  one  side  of  a  spherical  triangle  is  less  than 
the  sum  of  the  other  two. 

Let  AB  C  be  any  spherical  triangle ; 
then  any  side,  as  A  B,  is  less  than  the 
sum  of  the  other  two  sides,  A  C,  B  C. 

For,  draw  the  radii  OA,  OB,  OC, 
and  the  plane  angles  A  0  B,  A  0  C, 
COB  will  form  a  triedral  angle,  0. 
The  angles  A  OB,  AOC,  COB  will 
be  measured  by  AB,  AC,  B  C,  the 
side  of  the  spherical  triangle.  But  each  of  the  tlu^ce  plane 
angles  forming  a  triedral  angle  is  less  than  the  sum  of  the 
other  two  (Prop.  XIX.  Bk.  YII.).  Hence,  any  side  of  a 
spherical  triangle  is  less  than  the  sum  of  the  other  two. 

Proposition  III.  —  Theorem. 

523.  The  shortest  path  from  one  point  to  another^  on 
the  surface  of  a  sphere^  is  the  arc  of  the  great  circle  luhich 
joins  the  tivo  given  points. 

Let  ABD  be  the  arc  of  the 
great  circle  which  joins  tlie  points     ^ 
A  and  D  ;  then  the  line  A  B  D  is 
the  shortest  path  from  A  to  D  on 
the  surface  of  the  sphere. 

For,  if  possible,  let  the  shortest 
path  on  the  surface  from  A  to  D  pass  through  the  point 
C,  out  of  the  arc  of  the  great  circle  ABD.  Draw  A  C, 
D  C,  arcs  of  great  circles,  and  take  D  B  equal  to  D  C. 
Then  in  the  spherical  triangle  A  B  D  C  the  side  A  B  D  is 
less  than  tlie  sum  of  the  sides  AC,  D  C^  (Prop.  II.)  ;  and 


BOOK  IX.  219 

subtracting  the  equal  D  B  and  D  C,  there  will  remain  AB 
less  tlian  AC. 

Now,  the  shortest  path,  on  the  surface,  from  D  to  C, 
whether  it  is  the  arc  D  C,  or  any  other  line,  is  equal  to 
the  shortest  path  from  D  to  B  ;  for,  revolving  J)  C  about 
the  diameter  which  passes  through  D,  the  point  C  may  be 
brouglit  into  the  position  of  the  point  B,  and  the  shortest 
path  from  D  to  C  be  made  to  coincide  with  the  shortest 
path  from  D  to  B.  But,  by  hypothesis,  the  shortest  path 
from  A  to  D  passes  through  C  ;  consequently,  the  shortest 
path  on  the  surface  from  A  to  C  cannot  be  greater  than 
that  from  A  to  B. 

Now,  since  AB  has  been  proved  to  be  less  than  AC, 
the  shortest  path  from  A  to  C  must  be  greater  tlian  that 
from  A  to  B ;  but  this  has  just  been  shown  to  be  impos- 
sible. Hence,  no  point  of  the  shortest  path  from  A  to  D 
can  lie  out  of  the  arc  A  B  D  ;  consequently,  this  arc  of  a 
great  circle  is  itself  the  shortest  path  between  its  extrem- 
ities: 

524.  Cor.  The  distance  between  any  two  points  of  sur- 
face, on  the  surface  of  a  sphere,  is  measured  by  the  arc  of 
a  great  circle  joining  the  two  points. 

Proposition  TV.  —  Theorem. 

525.  The  sum  of  all  the  sides  of  any  spherical  polygon 
is  less  than  the  circumference  of  a  great  circle. 

Let  A  B  C  D  E  be  a  spherical  polygon  ;  E 

then  the  sum  of  the  sides  AB,  B  C,  CD,  /  I^^^D 

<fec.  is  less  than  the  circumference  of  a  A^      j     /j 

great  circle.  \  \B|Tc/ 

For,  from  0,  the  centre  of  the  sphere,  \\\  I 

draw  the  radii  OA,  OB,  OC,  <fec.,  and  \\|// 

the   plane   angles   A  0  B,  B  0  C,  COB,  M  li' 

&Q,.  will  form  a  polyedral   angle   at  0.  Y 

Now,  the  sum  of  the  plane  angles  which  ^ 


220 


ELEMENTS   OF   GEOMETRY. 


form  a  poljcdral  angle  is  less  than  four  right  angles 
(Prop.  XX.  Bk.  YIL).  Hence,  the  sum  of  the  arcs  AB, 
B  C,  CD,  &G.,  which  measure  these  angles,  and  bound 
the  spherical  polygon,  is  less  than  the  circumference  of  a 
great  circle. 

"526.  Cor.  The  sum  of  the  three  sides  of  a  spherical  tri- 
angle is  less  than  the  circumferejice  of  a  great  circle,  since 
a  triangle  is  a  polygon  of  three  sides. 


Proposition  Y.  —  Theorem. 

527.  The  extremities  of  a  diameter  of  a  sphere  are  the 
poles  of  all  circles  of  the  sphere  whose  planes  are  j)erpen- 
dicular  to  that  diameter. 

Let  D  E  be  a  diameter  per- 
pendicular to  AHB,  a  great 
circle  of  a  sphere,  and  also  to 
tlie  small  circle  FIG;  then 
D  and  E,  the  extremities  of 
this  diameter,  are  the  poles  of 
these  two  circles. 

For,  since  D  E  is  perpendic- 
ular to  the  plane  AHB,  it  is 
perpendicular  to  all  the  straight 
lines,  AC,  H  C,  B  C,  &c.,  drawn  through  its  foot  in  this 
plane  ;  hence,  all  the  arcs  D  A,  D  H,  D  B,  <fec.  are  quar- 
ters of  the  circumference.  So,  likewise,  are  all  the  arcs 
E  A,  EH,  E  B,  <fec. ;  hence  the  points  D  and  E  are  each 
equally  distant  from  all  the  points  of  the  circumference, 
AHB;  consequently  D  and  E  are  poles  of  that  circum- 
ference (Art.  501). 

Again,  since  the  i-adius  I)  C  is  perpendicular  to  the  plane 
AHB,  it  is  perpendicular  to  the  parallel  plane  FIG; 
hence  it  passes  through  0,  the  centre  of  the  circle  FIG 
(Prop.  I.  Cor.  4).  Hence,  if  the  oblique  lines  D  F,  D  I, 
D  G,  &c.  be  drawn,  these  lines  will  be  equally  distant  from 


BOOK  IX.  221 

the  perpendicular  D  0,  and  will  themselves  he  equal 
(Prop.  V.  Bk.  VII.).  But  the  chords  heing  equal,  the 
arcs  are  equal ;  hence  the  point  D  is  a  pole  of  the  small 
circle  FIG;  and,  for  like  reasons,  the  point  E  is  tlie 
other  pole. 

528.  Cor.  1.  Every  arc  of  a  great  circle,  D II,  drawn 
from  a  point  in  the  arc  of  a  great  circle,  AH'B,  to  its  pole, 
is  a  quarter  of  tl^e  circumference,  and  is  called  a  quad- 
rant. This  quadrant  makes  a  riglit  angle  with  the  arc 
AH.  For,  the  line  D  C  heing  perpendicular  to  the  plane 
AH  C,  every  plane  B  H  C  passing  through  the  line  D  C  is 
perpendicular  to  the  plane  A  H  C  (Prop.  YII.  Bk.  YII.)  ; 
hence  the  angle  of  those  planes,  or  the  angle  AHD,  is  a 
right  angle  (Art.  50G). 

529.  Cor.  2.  To  find  the  pole  of  a  given  arc,  AH,  draw 
the  indefinite  arc  HD  perpendicular  to  AH,  and  take  HD 
equal  to  a  quadrant ;  the  point  D  will  he  one  of  the  poles 
of  the  arc  AHD  ;  or  at  each  of  the  two  points  A  and  H, 
draw  the  arcs  AD  and  HD  perpendicular  to  AH;  the 
point  of  their  intersection,  D,  will  be  the  pole  required. 

530.  Cor.  3.  Conversely,  if  the  distance  of  the  point  D 
from  each  of  tlie  points  A  and  H  is  equal  to  a  quadrant, 
the  point  D  will  be  the  pole  of  the  arc  A  H  ;  and  the  an- 
gles D  A  H,  A  H  D  will  be  right. 

For,  let  C  be  the  centre  of  the  sphere,  and  draw  the 
radii  C  A,  C  D,  C  H.  Since  the  angles  A  C  D,  H  C  D  are 
right,  the  line  C  D  is  perpendicular  to  the  two  straight 
lines  C  A,  C  H  ;  hence  it  is  perpendicular  to  their  plane 
(Prop.  IV.  Bk.  VII.).  Hence  the  point  D  is  the  pole  of 
the  arc  A  H  ;  and  consequently  the  angles  D  A  H,  Jl  II D 
are  right  angles. 

531.  Scholium.  A  circle  may  be  described  on  the  surface 
of  a  sphere  with  the  same  facility  as  on  a  plane  surface. 
For  instance,  by  turning  the  arc  D  F,  or  any  other  lino 
extending  to  the  same  distance,  round  the  point  D    tho 


222  ELEMENTS  OF  GEOMETRY. 

extremity,  F,  will  describe  the  small  circle  FIG;  and  by 
turning  the  quadrant  D  F  A  round  the  point  D,  its  ex- 
tremity, A,  will  describe  the  great  circle  A  H  B. 

Proposition  VI.  —  Theorem. 

532.  A  plane  perpendicular  to  a  radius^  at  its  terminal 
Hon,  in  the  surface^  is  tangent  to  the  sphere. 

Let  A  D  B  be  a  plane  per-     a         *        D         E         B 
pendicular  to  a  radius,  C  D,  at 
its  termination,  D  ;  tlien  the 
plane  A  D  B  is  a  tangent  to 
the  sphere. 

For,  draw  from  the  centre, 
C,  any  other  straight  line,  CE, 
to  the  plane,  ADB.  Then, 
since  C  D  is  perpendicular  to 
the  plane,  it  is  shorter  than 
the  oblique  line  C  E  ;  hence  the  radius  C  F  is  shorter  than 
C  E  ;  consequently  the  point  E  is  without  the  sphere. 
The  same  may  be  shown  of  any  other  point  in  the  plane 
ADB,  except  the  point  D  ;  hence  the  plane  can  meet  the 
sphere  in  but  one  point,  and  therefore  is  a  tangent  to  the 
sphere  (Art.  503). 

533.  Scholimu.  In  the  same  manner,  it  may  be  proved 
that  two  spheres  ^re  tangent  to  each  other,  wlien  the  dis- 
tance between  their  centres  is  equal  to  the  sum  or  the  dif- 
ference of  their  radii ;  in  which  case  the  centres  and  the 
point  of  contact  lie  in  the  same  straight  line. 

Proposition  YII.  —  Theorem. 

534.  The  angle  formed  by  two  arcs  of  great  circles  is 
equal  to  the  angle  formed  by  the  tangents  of  those  arcs  at 
the  point  of  their  intersection,  and  is  measured  by  the  arc 
of  a  great  circle  described  from  its  vertex  as  a  pole,  and 
intercepted  between  its  sides,  produced  if  nScessary, 


BOOK   IX. 


223 


Let  B  A  C  be  an  angle  formed 
by  the  two  arcs  AB,  AC  ;  then 
will  it  be  equal  to  the  angle 
E  A  F,  formed  by  the  tangents 
A  E,  A  F,  and  it  is  measured 
by  B  C,  the  arc  of  a  great  circle 
described  from  the  vertex  A  as 
a  pole. 

For  the  tangent  AE,  drawn 
in  the  plane  of  the  arc  A  B,  is 
perpendicular  to  the  radius  A  0  (Prop.  X.  Bk.  III.)  ;  and 
the  tangent  A  F,  drawn  in  the  plane  of  the  arc  A  C,  is  per- 
pendicular to  the  same  radius  AO.  Hence  the  angle 
E  A  F  is  equal  to  the  angle  of  the  planes  A  0  B,  A  0  C 
(Art.  391)  ;  which  is  that  of  the  arcs  AB,  AC. 

Also,  if  the  arcs  A  B,  A  C  are  both  quadrants,  the  lines 
0  B,  0  C  will  be  perpendicular  to  A  0,  and  the  angle 
BOC  will  be  equal  to  the  angle  of  the  planes  A  OB,  AOC; 
hence  the  arc  B  C  is  the  measure  of  tlie  angle  of  tliese 
planes,  or  the  measure  of  the  angle  CAB. 

535.  Cor.  1.  The  angles  of  spherical  triangles  may  be 
compared  together,  by  means  of  the  arcs  of  great  circles 
described  from  their  vertices  as  poles,  and  included  be- 
tween their  sides  ;  hence  it  is  easy  to  make  an  angle  of 
this  kind  equal  to  a  given  angle. 

536.  Cor.  2.  Vertical  angles, 
such  as  A 0  C  and  BOD,  are 
equal  ;  for  each  of  them  is 
equal  to  the  angle  formed  by 
the  two  planes  A 0  B,  COD. 

It  is  also  evident  that  the 
two  adjacent  angles,  AOC, 
COB,  taken  together,  are 
equal  to  two  right  angles. 


224 


ELEMENTS    OF   GEOMETRY. 


Proposition  YIII.  —  Theorem. 

637.  If  from  the  vertices  of  any  spherical  triangle,  as 
poles,  arcs  of  great  circles  are  described,  a  second  trian- 
gle is  formed,  whose  vertices  will  he  poles  to  the  sides  of 
the  first  triangle. 

Let  A  B  C  be  any  spher- 
ical triangle ;  and  from 
the  vertices,  A,  B,  C,  as 
poles,  let  the  arcs  E  F, 
FD,  DE  be  described, 
and  a  second  triangle, 
D  E  F,  is  formed,  whose 
vertices,  D,  E,  F,  will  be 
poles  to  the  sides  of  the 
triangle  ABC. 

For,  the  point  A  being  the  pole  of  the  arc  E  F,  the  dis- 
tance AE  is  a  quadrant;  the  point  C  being  the  pole  of  tlie 
arc  D  E,  the  distance  C  E  is  also  a  quadrant ;  hence  tlie 
point  E  is  at  the  distance  of  a  quadrant  from  each  of  the 
points  A  and  C ;  hence  it  is  the  pole  of  the  arc  A  C  (Prop. 
V.  Cor.  3).  In  like  manner,  it  may  be  shown  that  D  is 
the  pole  of  the  arc  B  C,  and  F  that  of  the  arc  A  B. 

538.  Scholium.  Hence  the  triangle  ABC  may  be  de- 
scribed by  moans  of  D  E  F,  as  D  E  F  may  be  by  means  of 
ABC.  Spherical  triangles  tlius  described  are  said  to  l)e 
polar  to  each  other ^  and  are  called  polar  or  svpplcmental 
triangles. 

Proposition  IX. — Theorem. 

539.  Each  of  the  angles  of  a  spherical  tnangle  is  mcas- 
vred  by  a  semi-circumference  minus  the  side  lying  opposite 
to  it  in  the  polar  triangle. 

Let  A  B  C  be  a  spherical  triangle,  and  D  E  F  a  triangle 
polar  to  it ;  then  each  of  the  angles  of  A  B  C  is  measured 


BOOK  IX. 


225 


by  a  semi-circumference 
minus  the  side  lying  oppo- 
site to  it  in  D  E  F. 

For,  produce  the  sides 
A  B,  A  C,  if  necessary,  till 
they  meet  EF  in  G  and 
II.  The  point  A  being  the 
pole  of  the  arc  GH,  the 
angle  A  will  be  measured 
by  that  arc  (Prop.  VII.). 
But,  E  being  the  pole  of  A  H,  the  arc  E  H  is  a  quadrant ; 
and  F  being  the  pole  of  A  G,  F  G  is  a  quadrant.  Hence, 
E  H  and  G  F  together  are  equal  to  a  semi-circumference. 
Now,  the  sum  of  E  H  and  G  F  is  equal  to  the  sum  of  E  F 
and  G  H  ;  hence  the  arc  G  H,  which  measures  the  angle 
A,  is  equal  to  a  semi-circumference  minus  the  side  E  F. 
In  like  manner,  the  angle  B  will  be  measured  by  a  semi- 
circumference  minus  D  F  ;  and  the  angle  C  by  a  semi- 
circumference  minus  D  E. 

540.  Cor.  This  property  must  be  reciprocal  in  the  two 
triangles,  since  they  are  polar  to  each  other.  The  angle 
D,  for  example,  of  the  triangle  D  E  F,  is  measured  by  the 
arc  I K ;  but  the  sum  of  1 K  and  B  C  is  equal  to  the  sum 
of  I  C  and  B  K,  which  is  equal  to  a  semi-circumference  ; 
licnce  the  arc  IK,  the  measure  of  D,  is  equal  to  a  semi- 
circumference  minus  B  C.  In  like  manner,  it  may  be 
shown  tliat  E  is  measured  by  a  semi-circumference  miuus 
A  C,  and  F  by  a  semi-circumference  minus  A  B. 


Proposition  X. — Theorem. 

541.  Tlie  sum  of  the  ang-les  in  any  spherical  triangles 
is  less  than  six  right  angles^  and  greater  than  two. 

First.  Every  angle  of  a  spherical  triangle  is  less  than 
two  right  angles  ;  hence,  the  sum  of  the  three  is  less  than 
bix  right  angles. 


226  ELEMENTS  OF  GEOMETRY. 

Secondly.  The  measure  of  each  angle  of  a  spherical  tri- 
angle is  equal  to  the  semi-circumference  minus  the  corre- 
sponding side  of  the  polar  triangle  (Prop.  IX.)  ;  hence, 
the  sum  of  the  three  is  measured  hy  three  semi-circumfer- 
ences minus  the  sum  of  the  sides  of  the  polar  triangle. 
Now,  this  latter  sum  is  less  than  a  circumference  (Prop. 
lY.  Cor.)  ;  therefore,  taking  it  away  from  three  semi- 
circumferences,  the  remainder  will  bo  greater  than  ono 
semi-circumference,  which  is  the  measure  of  two  right 
angles  ;  hence,  the  sum  of  the  three  angles  of  a  spherical 
triangle  is  greater  than  two  right  angles. 

542.  Cor.  1.  The  sum  of  the  angles  of  ^  spherical  tri- 
angle is  not  constant,  like  that  of  the  angles  of  a  rectilineal 
triangle.  It  varies  between  two  right  angles  and  six, 
without  ever  arriving  at  either  of  these  limits.  Two 
given  angles,  therefore,  do  not  serve  to  determine  tlie 
third. 

543.  Cor.  2.  A  spherical  triangle  may  have  two,  or 
even  three  right  angles,  or  obtuse  angles. 

544.  Scholium.  If  a  spherical  triangle  has  two  riglit 
angles,  it  is  said  to  be  bi-rectang-ular ;  and  if  it  has  thi-ee 
right  angles,  it  is  said  to  be  tri-rectang-ular,  or  quadrantal. 
The  quadrantal  triangle  is  evidently  contained  eight  times 
in  the  surface  of  the  sphere. 

Proposition  XI.  —  Theorem. 

545.  If  around  the  vertices  of  any  two  ang-fes  of  a  i^iven 
spherical  triangle,  as  poles,  the  circumferences  of  two  cir- 
cles be  described,  which  shall  pass  through  the  third  angle 
of  the  triangle,  and  then  if  through  the  other  point  in 
vjhich  these  circmnferences  intersect,  and  the  vertices  of 
the  first  two  angles  of  the  triangles,  arcs  of  tivo  great 
circles  be  drawn,  the  triangle  thus  formed  ivill  liave  all  its 
parts  equal  to  those  of  the  given  triangle,  each  to  each. 


B 


BOOK  IX.  227 

Let  A  B  C  be  the  given  spherical 
triangle,  and  CED,  DFC  arcs  de- 
scribed about  the  vertices  of  any 
two  of  its  angles,  A  and  B,  as  poles; 
then  will  the  triangle  A  D  B  have 
all  its  parts  equal  to  those  of  ABC. 

For,  by  construction,  the  side  AD 
is  equal  to  AC,  D B  is  equal  to  BC, 
and  A  B  is  common  ;  hence  the  two 
triangles  have  tlieir  sides  equal,  each  to  each.  We  are 
now  to  show  that  the  angles  opposite  these  equal  sides  are 
also  equal. 

If  the  centre  of  the  sphere  is  supposed  to  be  at  0,  a  tri- 
edral  angle  may  be  conceived  as  formed  at  O  by  the  three 
plane  angles  A  OB,  AOC,  BOC;  also,  another  triedral 
angle  may  be  conceived  as  formed  by  the  three  plane  an- 
gles A  OB,  AOD,  BOD.  Now,  since  the  sides  of  the 
triangle  ABC  are  equal  to  those  of  the  triangle  A  D  B, 
the  plane  angles  forming  the  one  of  these  triedral  angles 
are  equal  to  the  plane  angles  forming  the  other,  each  to 
each.  Therefore  the  planes,  in  which  the  equal  angles 
lie,  are  equally  inclined  to  each  other  (Prop.  XXI.  Bk. 
VII.)  ;  hence,  all  the  angles  of  the  spherical  triangle 
DAB  are  respectively  equal  to  those  of  the  triangle 
CAB;  namely,  DAB  is  equal  to  B  AC,  DBA  to  ABC, 
and  A  D  B  to  A  C  B  ;  hence,  the  sides  and  angles  of  the 
triangle  A  D  B  are  equal  to  the  sides  and  the  angles  of  the 
triangle  A  C  B,  each  to  each. 

546.  Scholium.  The  equality  of  these  triangles  is  not, 
however,  an  absolute  equality,  or  one  of  superposition  ; 
for  it  would  be  impossible  to  apply  them  to  each  other 
exactly,  unless  they  were  isosceles.  The  equality  here 
meant  is  that  by  symmetry ;  therefore  the  triangles  AC  B, 
A  D  B  are  termed  symmetrical  triangles. 


228 


ELEMENTS  OF   GEOMETRY. 


Proposition  XII. — Theorem. 

547.  If  two  triangles  on  the  same  sphere^  or  on  equal 
spheres^  are  mutual///  equilateral,  the//  are  mutually  equi- 
angular;  and  their  equal  angles  are  opposite  to  equal  sides. 

Let  A  B  C,  A  B  D  be  two  triangles  on  j^ 

tlie  same  sphere,  or  on  eqvial  spheres, 
having  the  sides  of  the  one  respective- 
ly equal  to  those  of  the  other ;  then 
the  angles  opposite  to  the  equal  sides, 
in  the  two  triangles,  are  equal. 

For,  with  three  given  sides,  A  B, 
AC,  B C,  there  can  be  constructed 
only  two  trianglcj?,  A  C  B,  A  B  D,  and 
these  triangles  will  be  equal,  each  to 
each,  in  the  magnitude  of  all  their  parts  (Prop.  XI.). 
Hence,  these  two  triangles,  which  are  mutually  equilat- 
eral, must  be  either  absolutely  equal,  or  equal  by  symme- 
try; in  either  case  tliey  are  mutually  equiangular,  and  the 
equal  angles  lie  opposite  to  equal  sides. 


Proposition  XIII.  —  Theorem. 

548.  If  two  triangles  on  the  same  sphere,  or  on  equal 
spheres,  are  mutually  equiangular,  they  are  mutually  equi- 
lateral. 

Let  A  and  B  be  the  two  given  triangles  ;  P  and  Q,  their 
polar  triangles. 

Since  the  angles  are  equal  in  the  triangles  A  and  B,  the 
sides  will  be  equal  in  the  polar  triangles  P  and  Q  (Prop. 
IX.).  But  since  the  triangles  P  and  Q  are  mutually  equi- 
lateral, they  jnust  also  be  mutually  equiangular  (Prop. 
XII.)  ;  and,  tlio  angles  being  equal  in  the  triangles  P  and 
Q,  it  follows  that  the  sides  are  equal  in  their  polar  trian- 
gles A  and  B.     Hence,  the  triangles  A  and  B,  which  are 


BOOK   IX. 


229 


mutually  equiangular,  are  at   the   same   time  mutually 
equilateral. 

Proposition  XIY.  —  Theorem. 

549.  If  hvo  triangles  on  the  same  sphere,  or  on  equal 
spheres,  have  tvm  sides  and  the  included  angle  in  the  one 
equal  to  tivo  sides  and  the  included  angle  in  the  other ^ 
each  to  each,  the  tivo  triangles  are  equal  in  all  their  parts. 

lu  the  two  triangles  A. 
ABC,  DEF,  let  the 
side  A  B  be  equal  to  the 
side  D  E,  the  side  A  C 
to  the  side  D  F ;  and 
the  angle  B  A  C  to  the 
angle  E  D  F  ;  then  the 
triangles  will  he  equal 
in  all  their  parts.  B  E 

Let  the  triangle  D  E  G  be  symmetrical  with  the  triangle 
DEF  (Prop.  XL  Sch.),  having  the  side  E  G  equal  to  E  F, 
the  side  G  D  equal  to  F  D,  and  the  side  E  D  common,  and 
consequently  the  angles  of  the  one  equal  to  those  of  the 
other  (Prop.  XIL). 

Now,  the  triangle  ABC  may  be  applied  to  the  triangle 
D  E  F,  or  to  D  E  G  symmetrical  with  DEF,  just  as  two 
rectilineal  triangles  are  applied  to  each  other,  when  they 
have  an  equal  angle  included  between  equal  sides.  Heiice, 
all  the  parts  of  the  triangle  ABC  will  be  equal  to  all  the 
parts  of  the  triangle  D  E  ¥,  each  to  each  ;  that  is,  besides 
the  three  parts  equal  by  hypothesis,  we  shall  have  the  side 
B  C  equal  to  E  F,  the  angle  ABC  equal  to  DEF,  and 
the  angle  A  C  B  equal  to  D  F  E. 

550.  Cor.  If  two  triangles,  ABC,  DEF,  on  tlie  same 
sphere,  or  on  equal  spheres,  have  two  angles  and  the  in- 
cluded side  in  the  one  equal  to  two  angles  and  the  included 
side  in  the  other,  each  to  each,  the  two  triangles  are  equal 
in  all  their  parts. 

20 


230  ELEMENTS  OF  GEOMETRY. 

For  one  of  these  triangles,  or  the  triangle  symmetrical 
with  it,  may  be  applied  to  the  other,  as  is  done  in  the  cor- 
responding case  of  rectilineal  triangles. 

Proposition  XY.  —  Theorem. 

551.  In  every  isosceles  spherical  triangle,  the  angles  op> 
posite  the  equal  sides  are  equal ;  and,  conversely,  if  tivo 
angdes  of  a  spherical  triang-le  are  equal,  the  triangle  is 
isosceles. 

Let  ABC    be  an  isosceles  spherical  A 

triangle,  in  which  the  side  A  B  is  equal 
to  the  side  A  C  ;  then  will  the  angle  B 
be  equal  to  the  angle  C. 

For,  if  the  arc  A  D  be  drawn  from 
the«\^ertex  A  to  the  middle  point,  D,  of 
the  base, the  two  triangles  ABD,  ACD 
will  have  all  the  sides  of  the  one  re- 
spectively equal  to  the  corresponding  sides  of  tlie  other, 
namely,  A  D  common,  B  D  equal  to  D  C,  and  A  B  equal 
to  A  C  ;  hence  their  angles  must  be  equal ;  consequently, 
the  angles  B  and  C  are  equal. 

Conversely.  Let  the  angles  B  and  C  be  equal ;  then  will 
the  side  A  C  be  equal  to  A  B. 

For,  if  A  C  and  A  B  are  not  equal,  let  A  B  be  the  great- 
er of  the  two  ;  take  B  0  equal  to  A  C,  and  draw  0  C. 
The  two  sides  BO,  B  C  in  the  triangle  B  0  C  are  equal  to 
the  two  sides  A  B,  B  C  in  the  triangle  BAG;  the  angle 
0  B  C,  contained  by  the  first  two,  is  equal  to  A  C  B,  con- 
tained by  the  second  two.  Hence,  the  two  triangles  BOG, 
BAG  have  all  their  other  parts  equal  (Prop.  XIY. 
Gor.)  ;  hence  the  angle  0  G  B  is  equal  to  A  B  G.  But, 
by  hypothesis,  the  angle  A  B  G  is  equal  to  A  G  B  ;  hence 
we  have  0GB  equal  to  AG  B,  which  is  impossible  ;  there- 
fore A  B  cannot  be  unequal  to  A  C  ;  consequently  the 
sides  A  B,  AG,  opposite  the  equal  angles  B  and  G,  are 
equal. 


BOOK   IX.  231 

552.  Cor.  The  angle  B  A  D  is  equal  to  D  A  C,  and  the 
angle  B  D  A  is  equal  to  A  D  C  ;  the  last  two  are  therefore 
right  angles  ;  hence  the  arc  drawn  from  the  vertex  of  an 
isosceles  spherical  triangle  to  the  middle  of  the  base,  is 
perpendicular  to  the  base,  and  bisects  the  vertical  angle. 

Proposition  XVI. — Theorem. 

553.  In  a  spherical  triangle  ^  the  greater  side  is  opposite 
the.  greater  angle ;  and,  conversely,  the  greater  angle  is 
opposite  the  greater  side. 

In  the  triangle  ABC, 
let  the  angle  A  be  greater 
than  B  ;  then  will  the  side 
B  C,  opposite  to  A,  be 
greater  than  A  C,  opposite 
toB. 

Take  the  angle  BAD 
equal  to  the  angle  B ;  then, 
in  the  triangle  ABD,  we  shall  have  the  side  AD  equal 
to  DB  (Prop.  XV.).  But  the  sum  of  AD  plus  D  C  is 
greater  than  AC;  hence,  putting  DB  in  the  place  of  AD, 
we  shall  have  the  sum  of  D  B  plus  DC,  or  B  C,  greater 
than  AC. 

Conversely.  Let  the  side  B  C  be  greater  than  A  C  ;  then 
the  angle  B A C  will  be  greater  than  ABC.  For,  if  B A C 
were  equal  to  A  B  C,  we  should  have  B  C  equal  to  A  C  ; 
and  if  B  A  C  were  less  than  ABC,  we  sliould  then  have, 
as  has  just  been  shown,  B  C  less  than  A  C.  Both  of  these 
results  are  contrary  to  the  hypothesis  ;  hence  the  angle 
B  A  C  is  greater  than  ABC. 

Proposition  XVII. — Theorem. 

554.  If  two  triangles  on  the  same  sphere,  or  on  equal 
spheres,  are  mutually  equilateral,  they  are  equivalent. 


232 


ELEMENTS   OF   GEOMETRY. 


the  three  points  A,  B,  C ; 
and  they  will  all  be  equal 
At  the  point  F  make  the  angle  D  F  P 
make  the  arc  V  P  equal   to  C  0  ;  and 


Let  ABC,  DEFbe  two 
triangles,  having  the  three 
sides  of  the  one  equal  to  the 
three  sides  of  the  other,  each 
to  each,  namely,  AB  to  D  E, 
ACtoDF,  andCBtoEF; 
then  their  triangles  will  be 
equivalent. 

Let  0  be  the  pole  of  the 
small  circle   passing   through 
draw  the  arcs  0  A,  OB,  0  0 
(Prop.  V.  Sch.). 
equal   to   A  C  0  ; 
draw  DP,  E  P. 

The  sides  D  F,  F  P  are  equal  to  the  sides  A  C,  C  0,  and 
the  angle  D  F  P  is  equal  to  the  angle  A  0  0  ;  hence  the 
two  triangles  DFP,  AGO  are  equal  in  all  their  parts 
(Prop.  XIV.)  ;  hence  the  side  D  P  is  equal  to  AO,  and 
the  angle  D  P  F  is  equal  to  A  0  0. 

In  the  triangles  D  F  E,  ABC,  the  angles  D  FE,  ACB, 
opposite  to  the  equ^al  sides  D  E,  A  B,  are  equal  (Prop. 
XII.).  Taking  away  the  equal  angles  DFP,  A  C  0, 
there  will  remain  the  angle  PFE,  equal  to  0  C  B.  The 
sides  P  F,  F  E  are  'equal  to  the  .sides  0  C,  C  B  ;  hence  the 
two  triangles  F  P  E,  COB  are  equal  in  all  their  parts 
(Prop.  XIV.)  ;  hence  the  side  PE  is  equal  to  OB,  and 
the  angle  F  P  E  is  equal  to  C  0  B. 

Now,  the  triangles  D  F  P,  A  C  0,  which  have  the  sides 
equal,  each  to  each,  are  at  the  same  time  isosceles,  and 
may  be  applied  the  one  to  the  other.  For,  having  placed 
0  A  upon  its  equal  P  D,  the  side  0  C  will  fall  on  its  equal 
P  F,  and  thus  the  two  triangles  will  coincide  ;  conse- 
quently they  are  equal,  and  the  surface  D  P  F  is  equal 
to  A  0  C.  For  a  like  reason,  the  surface  F  P  E  is  equal 
to  C  0  B,  and  the  surface  D  P  E  is  equal  to  A  0  B  ;  hence 
we  have 


BOOK  IX. 


233 


AOC  +  COB  —  AOB  =  DPF  +  FPE  —  DPE, 
or,  A  B  C  =  D  E  F. 

Hence  the  two  triangles  A  B  C,  D  E  F  are  equivalent. 

555.  Cor.  1.  If  two  triangles  on  the  same  sphere,  or  on 
equal  spheres,  are  mutually  equiangular,  they  are  equiva- 
lent. For  in  that  case  the  triangles  will  be  mutually 
equilateral. 

556.  Cor.  2.  Hence,  also,  if  two  triangles  on  the  same 
sphere,  or  on  equal  spheres,  have  two  sides  and  the  includ- 
ed angle,  or  have  two  angles  and  the  included  side,  in  the 
one  equal  to  those  in  the  other,  the  two  triangles  are 
equivalent. 

557.  Scholium.  The  poles  0  and  P  might  lie  within  the 
triangles  A  B  C,  D  E  F  ;  in  which  case  it  would  be  requi- 
site to  add  the  three  triangles  DPF,  FPE,  DPE  together, 
to  form  the  triangle  D  E  F  ;  and  in  like  manner  to  add 
the  three  triangles  AOC,  COB,  AOB  together,  to  form 
the  triangle  ABC;  in  all  other  respects  the  demonstra- 
tion would  be  the  same. 


Peoposition  XYHI.  —  Theorem. 

558.  The  area  of  a  lune  is  to  the  surface  of  the  sphere 
as  the  angle  of  the  lune  is  to  four  right  angles^  or  as  the 
arc  which  measures  that  angle  is  to  the  circumference. 

Let  A  C  B  D  be  a  lune  upon 
a  sphere  whose  diameter  is 
A  B  ;  then  will  the  area  of  the 
lune  be  to  the  surface  of  the 
sphere  as  the  angle  D  0  C  to 
four  right  angles,  or  as  the  arc 
D  C  to  the  circumference  of  a 
great  circle. 

For,  suppose  the  arc  C  D  to 
be  to  tlie  circumference  C  D  E  F 
in  the  ratio  of  two  whole  numbers,  as  5  to  48,  for  example. 

20* 


234 


ELEMENTS   OP   GEOMETRY. 


Then,  if  the  circumference 
CDEF  be  divided  into  48 
equal  parts,  C  D  will  contain 
5  of  them ;  and  if  the  pole 
A  he  joined  with  the  several 
points  of  division  by  as  many 
quadrants,  we  shall  have  48 
triangles  on  the  surface  of  the 
hemisphere  ACDEF,  all  equal, 
since  all  their  parts  are  equal. 
Hence,  the  whole  sphere  must  contain  96  of  these  trian- 
gles, and  the  lune  A  C  B  D  10  of  them  ;  consequently,  the 
lune  is  to  the  sphere  as  10  is  to  96,  or  as  5  to  48 ;  that 
is,  as  the  arc  C  D  is  to  the  circumference. 

If  the  arc  C  D  is  not  commensurable  with  the  circum- 
ference, it  may  still  be  shown,  by  a  mode  of  reasoning  ex- 
emplified in  Prop.  XVI.  Bk.  III.,  that  the  lune  is  to  the 
sphere  as  C  D  is  to  the  circumference. 

559.  Cor.  1.  Two  lunes  on  the  same  sphere,  or  on 
equal  spheres,  are  to  each  other  as  the  angles  included 
between  their  planes. 

560.  Cor.  2.  It  has  been  shown  that  the  wiiole  surface 
of  the  sphere  is  equal  to  eight  quadrantal  triangles  (Prop. 
X.  Sell.).  Hence,  if  the  area  of  a  quadrantal  triangle  be 
represented  by  T,  the  surface  of  the  sphere  will  be  repre- 
sented by  8T.  Now,  if  the  right  angle  be  assumed  as 
unity,  and  the  angle  of  the  lune  be  represented  by  A,  we 

have, 

Area  of  the  lune  :  8  T  :  :  A  :  4, 

which  gives  the  area  of  lune  equal  to  2  A  X  T. 

561.  Cor.  3.  The  spherical  ungula  included  by  tlie 
planes  A  C  B,  A  D  B,  is  to  the  whole  sphere  as  the  angle 
D  O  C  is  to  four  right  angles.  For,  the  lunes  being  equal, 
the  spherical  ungulas  will  also  be  equal  ;  hence,  two 
spherical  ungulas  on  the  same  sphere,  or  on  equal  spheres, 


BOOK  IX.  235 

are  to  each  other  as  the  angles  included  between  their 
planes. 

Proposition  XIX.  —  Theorem. 

562.  If  two  great  circles  intersect  each  other  on  the  sur- 
face of  a  hemisphere^  the  sum  of  the  opposite  triangles 
thus  formed  is  equivalent  to  a  lune,  whose  angle  is  equal 
to  the  angle  formed  by  the  circles. 

Let  the  great  circles  BAD, 
C  A  E  intersect  on  the  surface 
of  a  hemisphere,  ABODE; 
then  will  the  sum  of  the  oppo- 
site triangles,  BAG,  DAE, 
be  equal  to  a  lune  whose  angle 
is  DAE. 

For,  produce  the  arcs  AD, 
A  E  till  they  meet  in  F  ;  and 
the  arcs  BAD,  ADF  will 
each  be  a  semi-circumference.  Now,  if  we  take  away 
A  D  from  both,  we  shall  have  D  F  equal  to  B  A.  For  a 
like  reason,  we  have  E  F  equal  to  C  A.  D  E  is  equal  to 
BC.  Hence,  the  two  triangles  BxlC,  DEF  are  mutually 
equilateral;  therefore  they  are  equivalent  (Prop.  XVII.). 
But  the  sum  of  the  triangles  DEF,  D  A  E  is  equivalent 
to  the  lune  A  D  F  E,  whose  angle  is  DAE. 

Proposition  XX.  —  Theorem. 

563.  The  area  of  a  spherical  triangle  is  equal  to  the 
excess  of  the  sum  of  its  three  angles  above  two  right  an- 
gles^ multiplied  by  the  quadrantal  triangle. 

Let  ABC  be  a  spherical  triangle  ;  its  area  is  equal  to 
tlie  excess  of  the  sum  of  its  angles,  A,  B,  C,  above  two 
right  angles  multiplied  by  the  quadrantal  triangle. 

For  produce  the  sides  of  the  triangle  ABC  till  they 


236 


ELEMENTS   OF   GEOMETRY. 


meet  the  great  circle  DEFGHI, 
drawn  without  the  triangle.  Tlie 
two  triangles  ADE,  AGH  are 
together  equivalent  to  the  lune 
whose  angle  is  A  (Prop.  XIX.), 
and  whose  area  is  expressed  by 
2  A  X  T  (Prop.  XVIII.  Cor.  2). 
Hence  we  have 

ADE  +  AGH  =  2AX  T; 

and,  for  a  like  reason, 

BGF+BID=2BxT,  andCIH  +  CFE=2CxT. 

But  the  sum  of  these  six  triangles  exceeds  the  hemisphere 
by  twice  the  triangle  ABC;  and  the  hemisphere  is  repre- 
sented by  4  T ;  consequently,  twice  the  triangle  A  B  C  is 
equivalent  to 

2AXT  +  2BXT  +  2CXT  — 4T; 

therefore,  once  the  triangle  A  B  C  is  equivalent  to 

(A  +  B  +  C  —  2)  X  T. 

Hence  the  area  of  a  spherical  triangle  is  equal  to  the  excess 
of  the  sum  of  its  three  angles  above  two  right  angles  mul- 
tiplied by  the  quadrantal  triangle. 

564.  Cor,  If  the  sum  of  the  three  angles  of  a  spherical 
triangle  is  equal  to  three  right  angles,  its  area  is  equal  to 
the  quadrantal  triangle,  or  to  an  eighth  part  of  the  suiface 
of  the  sphere  ;  if  the  sum  is  equal  to  four  right  angles,  the 
area  of  the  triangle  is  equal  to  two  quadrantal  triangles,  or 
to  a  fourth  part  of  the  surface  of  the  sphere,  <fec. 

Proposition  XXI. — Theorem. 

^^b.  The  area  of  a  spherical  polygon  is  equal  to  the 
excess  of  the  sum  of  all  its  angles  above  tivo  right  angles 
taken  as  many  times  as  the  polygon  has  sides,  less  twOy 
multiplied  by  the  quadrantal  triangle. 


BOOK  IX. 


237 


Let  ABODE  be  any  spherical 
polygon.  From  one  of  the  vertices, 
A,  draw  the  arcs  AC,  AD  to  the 
opjjosite  vertices ;  tlie  polygon  will 
be  divided  into  as  many  spherical 
triangles  as  it  has  sides  less  two. 
But  tlie  area  of  each  of  these  trian- 
gles is  equal  to  the  excess  of  the  sum  of  its  three  angles 
above  two  right  angles  multiplied  by  the  quadrantal  tri- 
angle (Prop.  XX.)  ;  and  the  sum  of  the  angles  in  all  the 
triangles  is  evidently  the  same  as  that  of  all  the  angles  in 
the  polygon  ;  hence  the  area  of  the  polygon  ABODE  is 
equal  to  the  excess  of  the  sum  of  all  its  angles  above  two 
right  angles  taken  as  many  times  as  the  polygon  has  sides, 
less  two,  multiplied  by  the  quadrantal  triangle. 

566.  Cor.  If  the  sum  of  all  the  angles  of  a  spherical 
polygon  be  denoted  by  S,  the  number  of  sides  by  w,  the 
quadrantal  triangle  by  T,  and  the  right  angle  be  regarded 
as  unity ^  the  area  of  the  polygon  will  be  expressed  by 


2  (n  — 2)  X  T=  (S-^2/j  +  4)  X  T, 


BOOK  X 


THE  THREE  ROUND  BODIES. 


DEFINITIONS. 


567.  A  Cylinder  is  a  solid,  which  may 
be  described  by  the  revolution  of  a  rectan- 
gle turning  about  one  of  its  sides,  which 
remains  immovable  ;  as  the  solid  described 
by  the  rectangle  A  B  C  D  revolving  about 
its  side  A  B. 

The  BASES  of  the  cylinder  are  the  circles 
described  by  the  sides,   AC,  BD,  of  the 
revolving  rectangle,  which  are  adjacent  to  the  immovable 
side,  A  B. 

The  AXIS  of  the  cylinder  is  the  straight  line  joining  the 
centres  of  its  two  bases  ;  as  the  immovable  line  A  B. 

The  CONVEX  SURFACE  of  the  cylinder  is  described  by  the 
side  C  D  of  the  rectangle,  opposite  to  the  axis  A  B. 

568.  A  Cone  is  a  solid  which  may  be  ^ 
described  by  the  revolution  of  a  right- 
angled  triangle  turning  about  one  of  its 
perpendicular  sides,  which  remains  im- 
movable ;  as  the  solid  described  by  the 
right-angled  triangle  ABC  revolving 
about  its  perpendicular  side  A  B. 

The  BASE  of  the  cone  is  the  circle  de- 
scribed by  the  revolution  of  the  side 
B  C,  which  is  perpendicular  to  the  im- 
movable side. 


BOOK  X.  239 

The  CONVEX  SURFACE  of  a  cone  is  described  by  the  hy- 
potlienuse,  A  C,  of  the  revolving  triangle. 

The  VERTEX  of  tlie  cone  is  the  point  A,  where  the  hy- 
pothenuse  meets  the  immovable  side. 

The  AXIS  of  the  cone  is  the  straight  line  joining  the 
vertex  to  the  centre  of  the  base  ;  as  the  line  A  B. 

The  ALTITUDE  of  a  cone  is  a  line  drawn  from  the  vertex 
perpendicular  to  the  base ;  and  is  the  same  as  the  axis,  AB. 

The  SLANT  HEIGHT,  or  SIDE,  of  a  cone,  is  a  straight  line 
drawn  from  tlie  vertex  to  the  circumference  of  the  base  ; 
as  the  line  A  C. 

569.  The  frustum  of  a  cone  is  the 
part  of  a  cone  included  between  the 
base  and  a  plane  parallel  to  the  base  ; 
as  the  solid  CD -P. 

The  AXIS,  or  altitude,  of  the  frus- 
tum, is  the  perpendicular  line  A  B  in- 
cluded between  the  two  bases;  and  the 
SLANT  HEIGHT,  or  SIDE,  is  that  portion  of  the  slant  height 
of  the  cone  which  lies  between  the  bases  ;  as  F  C. 

570.  Similar  Cylinders,  or  Cones,  are  tliose  whose 
axes  are  to  each  other  as  the  radii,  or  diameters,  of  their 
bases. 

571.  Tlie  sphere,  cylinder,  and  cone  are  termed  the 
Three  Round  Bodies  of  elementary  Geometry. 

Proposition  I.  —  Theorem. 

572.  Tlie  convex  surface  of  a  cylinder  is  equal  to  the' 
circumference  of  its  base  multiplied  by  its  altitude. 

Let  ABCDEF-Gbe  a  cylinder,  whose  circumference 
is  the  circle  A  B  C  D  E  F,  and  whose  altitude  is  the  line 
A  G  ;  then  its  convex  surface  is  equal  to  A  B  C  D  E  F 
multiplied  by  AG. 


240 


ELEMENTS   OF   GEOMETRY. 


In  tliG  base  of  the  cylinder  inscribe 
any  regular  polygon,  A  B  C  D  E  F,  and 
on  this  polygon  construct  a  right  prism 
of  the  same  altitude  with  the  cylinder. 
The  prism  will  be  inscribed  in  the  con- 
vex surface  of  the  cylinder.  The  con- 
vex surface  of  this  prism  is  equal  to  the 
perimeter  of  its  base  multiplied  by  its 
altitude,  AG  (Prop.  I.  Bk.  VIII.).  B  ~    C 

Conceive  now  the  arcs  subtending  the  sides  of  the  poly- 
gon to  be  continually  bisected,  until  a  polygon  is  formed 
having  an  indefinite  number  of  sides ;  its  perimeter  will 
then  be  equal  to  the  circumference  of  the  circle  ABCDEF 
(Prop.  XII.  Cor.,  Bk.  VI.)  ;  and  thus  the  convex  surface 
of  the  prism  will  coincide  with  the  convex  surface  of  the 
cylinder.  But  the  convex  surface  of  the  prism  is  always 
equal  to  the  perimeter  of  its  base  multiplied  by  its  alti- 
tude ;  hence,  the  convex  surface  of  the  cylinder  is  equal 
to  the  circumference  of  its  base  multiplied  by  its  altitude. 

573.  Cor.  1.  If  two  cylinders  have  the  same  altitude, 
their  convex  surfaces  are  to  each  other  as  the  circumfer- 
ences of  their  bases. 

574.  Cor.  2.  If  H  represent  the  altitude  of  a  cylinder, 
and  R  the  radius  of  its  base,  then  we  shall  have  the  cir- 
cumference of  the  base  represented  by  2R  X  n  (Prop. 
XV.  Cor.  3,  Bk.  VI.),  and  the  convex  surface  of  the  cyl- 
inder by  2  R  X  TT  X  H. 


Proposition  II.  —  Theorem. 

575.  The  solid  contents  of  a  cylinder  are  equal  to  the 
product  of  its  base  by  its  altitude. 

Let  ABCDEF-Gr  be  a  cylinder  whose  base  is  the 
circle  ABCDEF,  and  whoso  altitude  is  the  line  A  G ; 
then  its  solid  contents  are  equal  to  the  product  of 
ABCDEF  by  AG. 


BOOK  X. 


241 


/F 


£ 


In  the  base  of  the  cylhider  inscribe 
any  regular  polygon,  A  B  C  D  E  F,  and 
on  this  polygon  construct  a  right  prism 
of  the  same  altitude  with  the  cylinder. 
The  prism  will  be  inscribed  in  the  con- 
vex surface  of  the  cylinder.  The  solid 
contents  of  this  prism  are  equal  to 
the  product  of  its  base  by  its  altitude 
(Prop.  XIII.  Bk.  VIIL).  F~^C 

Conceive  now  the  number  of  the  sides  of  the  polygon  to 
be  indefinitely  increased,  until  its  perimeter  coincides  with 
the  circumference  of  the  circle  A  B  C  D  E  F  (Prop.  XII. 
Cor.,  Bk.  VI.),  and  the  solid  contents  of  the  prism  will 
equal  those  of  the  cylinder.  But  the  solid  contents  of  the 
prism  will  still  be  equal  to  the  product  of  its  base  by  its 
altitude  ;  hence  the  solid  contents  of  the  cylinder  are 
equal  to  the  product  of  its  base  by  its  altitude. 

576.  Co?'.  1.  Cylinders  of  the  same  altitude  are  to  each 
other  as  their  bases ;  and  cylinders  of  equal  bases  are  to 
each  other  as  their  altitudes. 

577.  Cor.  2.  Similar  cylinders  are  to  each  other  as  the 
cubes  of  their  altitudes,  or  as  the  cubes  of  the  diameters 
of  their  bases.  For  the  bases  are  as  the  squares  of  their 
radii  (Prop.  XIII.  Bk.  VI.),  and  the  cylinders  being  simi- 
lar, the  radii  of  their  bases  are  to  each  other  as  their  alti- 
tudes (Art.  570)  ;  therefore  the  bases  are  as  the  squares 
of  the  altitudes  ;  hence,  the  products  of  the  bases  by  the 
altitudes,  or  the  cylinders  themselves,  are  as  the  cubes  of 
the  altitudes. 

578.  Cor.  3.  If  the  altitude  of  a  cylinder  be  represented 
by  H,  and  the  area  of  its  base  by  R*  X  ^  (Prop.  XV.  Cor. 
2,  Bk.  VI.),  the  solid  contents  of  the  cylinder  will  bo  rep- 
resented by  R-  X  ^  X  H. 

21 


242  ELEMENTS  OF  GEOMETRY. 


Proposition  III.  —  Theorem. 

579.  The  convex  surface  of  a  cone  is  equal  to  the  cir- 
cumference of  the  base  multiplied  h\j  half  the  slant  height. 

Let  A  B  C  D  E  F-S  be  a  cone  S 

whose  base  is  the  circle  A  B  C  D  E  F, 
and  whose  slant  height  is  the  line 
S  A  ;  then  its  convex  surface  is 
equal  to  A  B  C  D  E  F  multiplied  by 
I  SA. 


In  the  base  of  the  cone  inscribe 
any  regular  polygon,  A  B  C  D  E  F, 
and  on  this  polygon  construct  a  reg- 
ular pyramid  having  tlie  same  ver- 
tex, S,  with  the  cone.  Then  a  right  pyramid  will  be  in- 
scribed in  the  cone. 

From  S  draw  S  H  perpendicular  to  B  C,  a  side  of  the 
polygon.  The  convex  surface  of  tlie  pyramid  is  equal  to 
the  perimeter  of  its  base,  multiplied  by  half  its  slant 
height,  SH  (Prop.  XV.  Bk.  YIII.).  Conceive  now  the 
arcs  subtending  the  sides  of  the  polygon  to  be  continually 
bisected,  until  a  polygon  is  formed  having  an  indefinite 
number  of  sides  ;  its  perimeter  will  equal  the  circumfer- 
ence of  the  circle  A  B  C  D  E  F  ;  its  slant  height,  S  H,  will 
equal  that  of  the  cone,  and  its  convex  surface  coincide 
with  the  convex  surface  of  the  cone.  But  the  convex 
surface  of  every  right  pyramid  is  equal  to  the  perimeter 
of  its  base,  multiplied  by  half  the  slant  height ;  hence  the 
convex  surface  of  the  cone  is  equal  to  the  circumference 
of  its  base  multiplied  by  half  its  slant  height. 

580.  Cor.  If  S  A  represent  the  slant  height  of  a  cone, 
and  R  the  radius  of  the  base,  then,  since  the  circumference 
of  the  base  is  represented  by  2  R  X  ^r  (Prop.  XY.  Cor.  3, 
Bk.  YI.),  the  convex  surface  of  the  cone  will  be  repre- 
sented by  2  R  X  71  X  ^  S  A,  equal  to  tt  X  R  X  S  A. 


BOOK    X. 


243 


Proposition  IY.  —  Theorem. 

581.  The  convex  surface  of  a  frustum  of  a  cone  is  eqnal 
to  half  the  sum  of  the  circumference  of  the  two  bases 
multiplied  by  its  slant  height. 

Let  A  B  C  D  E  F  -  M  be  the  frustum 
of  a  cone,  and  AG  its  slant  height; 
tlieu  the  convex  surface  is  equal  to 
half  the  sum  of  the  circumferences  of 
the  two  bases  ABCDEF,  GHIKLM, 
multiplied  by  A  G.    . 

For,  inscribe  in  the  bases  of  the  frus- 
tum two  regular  polygons  of  tlie  same  ^  ^ 
number  of  sides,  having  their  sides  parallel,  each  to  each. 
Draw  the  straight  lines  AG,  B  H,  CI,  <fcc.,  joining  the 
vertices  of  the  corresponding  angles,  and  these  lines  will 
be  the  edges  of  the  frustum  of  a  pyramid  inscribed  in  the 
frustum  of  the  cone.  The  convex  surface  of  the  frustum 
of  the  pyramid  is^^qual  to  half  the  sum  of  the  perimeters 
of  the  two  bases  multiplied  by  its  slant  height,  ON  (Prop. 
XVII.  Bk.  VIII.) . 

Conceive  now  the  number  of  sides  of  the  inscribed  poly- 
gons to  be  indefinitely  increased  ;  the  perimeters  of  the 
polygons  will  then  coincide  with  the  circumferences  of  the 
circles  ABCDEF,  GHIKLM;  and  the  slant  height, 
ON,  of  the  frustum  of  the  pyramid,  will  equal  the  slant 
height,  A  G,  of  the  frustum  of  the  cone ;  and  the  surfaces 
of  the  two  frustums  will  coincide. 

But  the  convex  surface  of  every  frustum  of  a  right 
pyramid  is  equal  to  half  the  sum  of  the  perimeters  of  its 
two  bases,  multiplied  by  its  slant  height ;  hence,  the  con- 
vex surface  of  the  frustum  of  the  cone  is  equal  to  half  the 
sum  of  the  circumference  of  its  two  bases  multiplied  by 
half  its  slant  height. 

582.  Cor.  Through  R,  the  middle  point  of  the- side  KD, 


244 


ELEMENTS   OF   GEOMETRY. 


draw  the  diameter  RST,  parallel  to  the 

diameter  AQD,  and  the  straight  lines 

RU,  KY,   parallel   to   the  axis   P  Q. 

Then,  since  D  R  is  equal  to  R  K,  D  U 

is  equal  to  UY   (Prop.  XYII.  Cor.  2, 

Bk.  lY.)  ;  hence,  the  radius  S  R  is  equal 

to  half  the  sum  of  the  radii  QD,  PK. 

But  the  circumferences  of  circles  being  to  each  other  as 

their  radii  (Prop.  XIII.  Bk.  YI.),  the  circumference  of 

the  section  of  which  S  R  is  the  radius  is  equal  to  half  the 

sum  of  the  circumferences  of  which  QP,  PK  are  the 

radii ;  hence,  the  convex  surface  of  a  frustum  of  a  cone  is 

equal  to  the  slant  height  multiplied  by  the  circumference 

of  a  section  at  equal  distances  between  the  two  bases. 


Proposition  Y 


Theorem. 


583.  The  solidity  of  a  cone  is  equal  to  the  product  of  its 
base  by  one  third  of  its  altitude. 

Let  ABCDEF-S  be  a  cone, 
whose  base  is  A  B  C  D  E  F,  and  alti- 
tude S  II  ;  then  its  solidity  is  equal 
to  ABCDEF  X  ^SII. 

In  the  base  of  the  cone  inscribe 
any  regular  polygon,  ABCDEF, 
and  on  this  polygon  construct  a  reg- 
ular pyramid,  having  the  same  vertex, 
S,  with  the  cone.  Then  a  right  pyra- 
mid will  be  inscribed  in  the  cone  ; 
and  its  solidity  will  be  equal  to  the  product  of  its  base  by 
one  third  of  its  altitude  (Prop.  XX.  Bk.  YIIL). 

Conceive,  now,  the  number  of  sides  of  the  polygon  to  bo 
indefinitely  increased,  and  its  perimeter  will  become  equiil 
to  the  circumference  of  the  cone,  and  the  pyramid  will 
exactly  coincide  with  the  cone.  But  the  solidity  of  every 
right  pyramid  is  equal  to  the  product  of  the  base  by  one 


BOOK   X. 


245 


third  of  its  altitude  ;  licnce,  the  solidity  of  a  cone  is  equal 
to  the  product  of  its  base  by  one  third  of  its  altitude. 

584.  Cor.  1.  A  cone  is  the  third  of  a  cylinder  havuig 
the  same  base  and  the  same  altitude ;  hence  it  follows, — 

1.  That  cones  of  equal  altitudes  are  to  each  other  as 
their  bases ; 

2.  That  cones  of  equal  bases  are  to  each  other  as  their 
altitudes  ; 

3.  That  similar  cones  are  as  the  cubes  of  the  diameters 
of  their  bases,  or  as  the  cubes  of  their  altitudes. 

585.  Cor.  2.  If  the  altitude  of  a  cone  be  represented  by 
H,  and  the  radius  of  its  base  by  R,  the  solidity  of  the  cone 
will  be  represented  by 

R^  X  71  X  *  H,     or     i  71  X  R'  X  H. 


Proposition  YI.  —  Theorem. 

586.  The  solidity  of  the  frustum  of  a  cone  is  equivalent 
to  the  sum  of  three  cones ^  having-  for  their  common  alti- 
tude the  altitude  of  the  frustum,  and  lohose  bases  are  the 
two  bases  of  the  frustum,  and  a  mean  proportional  between 
them. 

Let  A  B  C  D  E  F  -  M  be  the  frur- 
turn  of  a  cone ;  then  will  its  solidity 
be  equivalent  to  the  sum  of  three 
cones  having  the  same  altitude  as 
the  frustum,  and  whose  bases  arc 
tlic  two  bases  of  the  frustum,  and  a 
mean  proportional  between  them. 

For,  inscribe  in  the  two  bases  of 
the  frustum  two  regular  polygons 
having  the  same  number  of  sides,  ^  ^ 

and  having  their  sides  parallel,  each  to  each.  Lot  the 
vertices  of  the  corresponding  angles  be  joined  by  the 
straight  lines  BII,  CI,  &c.,  and  there  is  inscribed  in  the 

21* 


246  ELEMENTS  OF  GEOMETRY. 

frustum  of  tlio  cone  the  frustum  of  a  regular  pyramid. 
The  solidity  of  the  frustum  of  this  pyramid  is  equivalent 
to  the  sum  of  three  pyramids,  having  for  their  common 
altitude  the  altitude  of  the  frustum,  and  whose  hases  are 
the  two  bases  of  the  frustum,  and  a  mean  proportional 
between  them. 

Conceive  now  the  number  of  the  sides  of  the  polygons 
to  be  indefinitely  increased  ;  and  the  bases  of  the  frustum 
of  the  pyramid  will  equal  the  bases  of  the  frustum  of  the 
cone  ;  and  the  two  frustums  will  coincide.  Hence  the 
frustum  of  a  cone  is  equivalent  to  the  sum  of  three  cones, 
having  for  their  common  altitude  the  altitude  of  the  frus- 
tum, and  whose  bases  are  the  two  bases  of  the  frustum, 
and  a  mean  proportional  between  them. 

Proposition  YII.  —  Theorem. 

587.  If  any  regular  semi-polygon  be  revolved  about  a 
line  passing  through  the  centre  and  the  vertices  of  oppo- 
site angles,  the  surface  described  will  be  equal  to  the  pro- 
duct of  its  axis  by  the  circumference  of  its  inscribed  circle. 

Let  the  regular  semi-polygon  A  B  C  D  E  F 
be  revolved  about  AF  as  an  axis;  tlicn 
the  surface  described  by  the  sides  AB, 
BC,  CD,  &c.  will  equal  the  product  of 
AF  by  the  inscribed  circle. 

For,  from  the  vertices  B,  C,  D,  E  of  the 
semi-polygon,  draw  B  G,  CH,  DM,  EN, 
perpendicular  to  the  axis  A  F  ;  and  from 
the  centre,  0,  draw  0 1  perpendicular  to 
one  of  the  sides;  also  draw  IK  perpendicular  to  AF,  anc] 
B  L  perpendicular  to  C  H. 

Now  0  I  is  the  radius  of  the  inscribed  circle  (Prop.  II. 
Bk.  VI.)  ;  and  the  surface  described  by  the  revolution  of 
a  side,  B  C,  of  a  regular  polygon,  is  equal  to  B  C  multiplied 
by  the  circumference,  IK  (Prop.  IV.  Cor.). 


BOOK  X.  247 

The  two  triangles  OIK,  B  C  L,  having  their  sides  per- 
pendicular to  each  other,  are  similar  (Prop.  XXV.  Bk. 
IV.)  ;  therefore, 

B  C  :  BL  or  GH  :  :  01  :  IK  :  :  Circ.Ol  :  Circ.  IK. 

Hence  (Prop.  I.  Bk.  II.), 

BC  X  Circ.  IK=  GH  X  Circ.  01; 

that  is,  the  surface  described  by  B  C  is  equal  to  the  pro- 
duct of  the  altitude  G  H  b j  the  circumference  of  the  in- 
scribed circle.  The  same  may  be  shown  of  each  of  the 
other  sides  ;  hence,  the  surface  described  by  all  the  sides 
taken  together  is  equal  to  the  product  of  the  sum  of  the 
altitudes  AG,  GH,  HM,  MN,  N  F,  by  the  circ.  0  I,  or 
to  the  product  of  the  axis  A  F  by  the  circ.  0 1. 

Proposition  VIII.  —  Theorem. 

588.  The  surface  of  a  sphere  is  equal  to  the  product  of 
its  diameter  by  the  circumference  of  a  great  circle. 

Let   ABCDEF   be  a   semicircle    in  x 

which  is  inscribed  any  regular  scmi-poly-  b^^#^ 

gon  ;  from  the  centre,  0,  draw  0 1  per-  ]U^ 

pendicular  to  one  of  the  sides.  C  r      ^-j." 

If  now  the  semicircle  and   the  semi- 
polygon  be  revolved  about  the  axis  A  F,        D  v~ 

tho  surface  described  by  the  semicircle  \v 

will  be  the  surface  of  a  sphere  (Art.  497),  ^^ 

and  that  described  by  the  semi-polygon  ^ 

will  be  equal  to  the  product  of  its  axis,  AF,  by  the  cir- 
cumference, 0 1  (Prop.  VII.)  ;  and  the  same  is  true, 
whatever  be  the  number  of  sides  of  the  polygon. 

Conceive  the  number  of  sides  of  the  semi-polygon  to 
be  made,  by  continual  bisections,  indefinitely  great ;  tlicu 
its  perimeter  will  coincide  with  the  semi-circumferenco 
ABCDEF,  and  the  perpendicular  0 1  will  be  equal  to 
the  radius  0  A ;  hence,  the  surface  of  the  sphere  is  equal 


248  ELEMENTS  OF  GEOMETRY. 

to  the  product  of  the  diameter  by  the 
circumference  of  a  great  circle.  ^^i^. 

589.  Cor.  1.    The  surface  of  a  sphere  //\ 
is  equal  to  the  area  of  four  of  its  great          I 
circles.                                                                   I 

For  the  area  of  a  circle  is  equal  to  the  -^V ^^ 

product  of  the  circumference  by  half  the  e^^^^ 

radius,   or  one   fourth   of  the   diameter  ^^^^^^F 

(Prop.  XY.  Bk.VI.). 

590.  Cor.  2.  The  surface  of  a  zone  or  seg-ment  is  equal 
to  the  product  of  its  altitude  hy  the  circumference  of  a 
great  circle. 

For  the  surface  described  by  the  sides  B  C,  C  I)  of  the 
inscribed  polygon  is  equal  to  the  product  of  the  altitude 
G  M  by  the  circumference  of  the  inscribed  circle  0 1.  If, 
now,  the  number  of  the  sides  of  an  inscribed  polygon  be 
indefinitely  increased,  its  perimeter  will  equal  the  circle, 
and  B  C,  C  D  will  coincide  with  tlie  arc  BCD;  conse- 
quently, the  surface  of  the  zone  described  by  the  revolution 
of  B  C  D  is  equal  to  the  product  of  its  altitude  by  the  cir- 
cumference of  a  great  circle.  In  like  manner,  the  same 
may  bo  proved  true  of  a  segment,  or  a  zone  having  but 
one  base. 

591.  Cor.  3.  The  surfaces  of  two'  zones,  or  segments 
upon  the  same  sphere,  are  to  eacli  other  as  their  altitudes  ; 
and  any  zone  or  segment  is  to  the  surface  of  the  sphere  as 
the  altitude  of  that  zone  or  segment  is  to  the  diameter. 

592.  Cor.  4.  If  the  radius  of  a  sphere  is  represented  by 
R,  and  its  diameter  by  D,its  surface  will  be  represented  by 

4  71  X  B-,     or     71  X  I)^ 

593.  Cor.  5.  Hence,  the  surfaces  of  spheres  are  to  each 
otlicr  as  the  squares  of  their  radii  or  diameters. 

594.  Cor.  6.   If  the  altitude  of  a  zone  or  segment  is 


BOOK  X.  249 

represented  by  H,  the  surface  of  a  zone  or  segment  will 
be  represented  by 

2  7t  X  R  X  H,     or     TT  X  D  X  H. 

Proposition  IX.  —  Theorem. 

595.  The  solidity  of  a  sphere  is  equal  to  the  product  of 
its  surface  by  one  third  of  its  radius. 

For  a  sphere  may  be  regarded  as  composed  of  an  indefi- 
nite number  of  pyramids,  each  having  for  its  bp-se  a  part 
of  the  surface  of  tlie  sphere,  and  for  its  vertex  the  centre 
of  the  sphere  ;  consequently,  all  these  pyramids  have  the 
radius  of  the  sphere  as  their  common  altitude. 

Now,  the  solidity  of  every  pyramid  is  equal  to  tlie  pro- 
duct of  its  base  by  one  third  of  its  altitude  (Prop.  XX. 
Bk.  VIII.);  hence,  the  sum  of  the  solidities  of  these  pyra- 
mids is  equal  to  the  product  of  tlie  sum  of  their  bases  by 
one  third  of  their  common  altitude.  But  the  sum  of  their 
bases  is  the  surface  of  the  spliere,  and  their  common  altitude 
its  radius  ;  consequently,  the  solidity  of  the  sphere  is  equal 
to  the  product  of  its  surface  by  one  third  of  its  radius. 

596.  Cor.  1.  The  solidity  of  a  spherical  pyramid  or 
sector  is  equal  to  the  product  of  the  polygon  or  zone  ivhich 
forms  its  base,  by  one  third  of  the  radius. 

For  the  polygon  or  zone  forming  the  base  of  the  spheri- 
cal pyramid  or  sector  may  be  regarded  as  composed  of  an 
indefinite  number  of  planes,  each  serving  as  a  base  to  a 
pyramid,  having  for  its  vertex  the  centre  of  the  sphere. 

597.  Cor.  2.  Spherical  pyramids,  or  sectors  of  the 
same  sphere  or  of  equal  spheres,  are  to  eacli  other  as  their 
bases. 

598.  Cor.  3.  A  spherical  pyramid  or  sector  is  to  tlie 
sphere  of  which  it  is  a  part,  as  its  base  is  to  the  surface  of 
the  sphere. 

599.  Cor.  4.   Hence,  spherical  sectors  upon  the  same 


250  ELEMENTS    OF   GEOMETRY. 

sphere  are  to  each  other  as  the  altitudes  of  the  zones  form- 
ing their  bases  (Prop.  YIII.  Cor.  3)  ;  and  any  spherical 
sector  is  to  the  sphere  as  the  altitude  of  the  zone  forming 
its  base  is  to  the  diameter  of  the  sphere. 

GOO.  Cor.  5.  If  the  radius  of  a  sphere  is  represented  by 
R,  its  diameter  by  D,  and  its  surface  by  S,  its  solidity  will 
be  represented  by 

SxiR  =  47rXR'X-^R==|^XR'ori7rXr)^ 

601.  Cor  6.  Hence,  the  solidities  of  spheres  are  to  each 
other  as  the  cubes  of  their  radii. 

602.  Cor.  7.  If  the  altitude  of  the  zone  which  forms 
the  base  of  a  sector  be  represented  by  H,  the  solidity  of 
the  sector  will  be  represented  by 

2  71  X  R  X  H  X  ^  B  =  §  ^  X  H'  X  H. 

603.  Scholium.  The  solidity  of  the  spher- 
ical segment  less  than  a  hemisphere,  and  of 
one  base,  formed  by  the  revolution  of  a  por- 
tion, AB  C,  of  a  semicircle  about  the  radius 
OA,  is  equivalent  to  the  solidity  of  the 
spherical  sector  formed  by  A  0  B,  less  the 
solidity  of  the  cone  formed  by  0  B  C. 

The  solidity  of  the  spherical  segment 
greater  than  a  hemisphere,  and  of  one  base,  formed  by  the 
revolution  of  ADE,  is  equivalent  to  the  solidity  of  the 
splierical  sector  formed  by  A  0  D,  plus  the  solidity  of  the 
cone  formed  by  0  D  E. 

The  solidity  of  the  spherical  segment  of  two  bases 
formed  by  the  revolution  of  C  B  D  E  about  the  axis  A  F, 
is  equivalent  to  the  solidity  of  the  segment  formed  by 
ADE,  less  the  solidity  of  the  segment  formed  by  ABC. 

Proposition  X. — Theorem. 

604.  The  surface  of  a  sphere  is  equivalent  to  the  coniiex 
surface  of  the  circumscribed  cylinder^  and  is  two  thirds 


BOOK   X. 


of  the  whole  surface  of  the  cylinder ;  also,  the  solidity  of 
the  spite  re  is  two  thirds  of  that  of  the  circumscribed  cyl- 
inder. 

Let  ABFI  be  a  great  circle  of 
the  sphere  ;  D  E  G  H  the  circum- 
scribed square  ;  then,  if  the  semi- 
circle A  B  F  and  the  semi-square 
A  D  E  F  be  revolved  about  the  di- 
ameter AF,  the  semicircle  will 
describe  a  sphere,  and  the  semi- 
square  a  cylinder  circumscribing 
the  sphere. 

The  convex  surface  of  the  cylinder  is  equal  to  the  cir- 
cumference of  its  base  multiplied  by  its  altitude  (Prop. 
I.).  But  the  base  of  the  cylinder  is  equal  to  the  great 
circle  of  the  sphere,  its  diameter  E  G  being  equal  to  the 
diameter  B  I,  and  the  altitude  D  E  is  equal  to  the  diam- 
eter AF  ;  hence,  the  convex  surface  of  the  cylinder  is 
equal  to  the  circumference  of  the  great  circle  multiplied 
by  its  diameter.  This  measure  is  the  same  as  that  of  the 
surface  of  the  sphere  (Prop.  VIII.)  ;  hence,  the  surface  of 
the  sphere  is  equal  to  the  convex  surface  of  the  circum- 
scribed cylinder. 

But  the  surface  of  the  sphere  is  equal  to  four  great  cir- 
cles of  the  sphere  (Prop.  VIII.  Cor.  1)  ;  hence,  the  convex 
surface  of  the  cylinder  is  also  equal  to  four  great  circles  ; 
and  adding  the  two  bases,  each  equal  to  a  great  circle, 
the  whole  surface  of  the  circumscribed  cylinder  is  equal 
to  six  great  circles  of  the  sphere ;  hence,  the  surface  of 
the  sphere  is  f  or  f  of  the  wliole  surface  of  the  circum- 
scribed sphere. 

In  the  next  place,  since  the  base  of  the  circumscribed 
cylinder  is  equal  to  a  great  circle  of  the  sphere,  and  its 
altitude  to  the  diameter,  the  solidity  of  the  cylinder  is 
equal  to  a  great  circle  multiplied  by  its  diameter  (Prop. 
II.).     But  the  solidity  of  the  splierc  is  equal  to  its  sur- 


252  ELEMENTS   OP   GEOMETRY. 

face,  or  four  great  circles,  multiplied  by  one  third  of  its 
radius  (Prop.  IX.),  which  is  the  same  as  one  great  circle 
multiplied  by  |  of  the  radius,  or  by  f  of  the  diameter ; 
hence,  the  solidity  of  the  sphere  is  equal  to  f  of  that  of 
the  circumscribed  cylinder. 

605.  Cor.  1.  Hence  the  sphere  is  to  the  circumscribed 
cylinder  as  2  to  3  ;  and  their  solidities  are  to  each  other 
as  their  surfaces. 

606.  Cor.  2.  Since  a  cone  is  one  third  of  a  cylinder  of 
the  same  base  and  altitude  (Prop.  V.  Cor.  1),  if  a  cone 
has  tlie  diameter  of  its  base  and  its  altitude  each  equal  to 
the  diameter  of  a  given  sphere,  the  solidities  of  the  cone 
and  sphere  are  to  each  other  as  1  to  2  ;  and  the  solidities 
of  the  cone,  sphere,  and  circumscribing  cylinder  are  to 
each  other,  respectively,  as  1,  2,  and  3. 


BOOK   XI. 

APPLICATIONS   OF    GEOMETRY    TO   THE    MENSU- 
RATION  OF   PLANE   FIGURES. 

DEFINITIONS. 

607.  Mensuration  of  Plane  Figures  is  the  process  of 
determining  the  areas  of  plane  surfaces. 

608.  The  area  of  a  figure,  or  its  quantity  of  surface,  is 
determined  by  the  number  of  times  the  given  surface  con- 
tains some  other  area,  assumed  as  the  unit  of  measure. 

609.  The  measuring  unit  assumed  for  a  given  surface 
is  called  the  svperficial  unit,  and  is  usually  a  square,  tak- 
ing its  name  from  the  linear  unit  forming  its  side  ;  as  a 
square  whose  side  is  1  inch,  1  foot,  1  yard,  &c. 

Some  superficial  units,  however,  have  no  corresponding 
linear  unit ;  as  the  rood,  acre,  &c. 


610. 

Table  of  Line. 

iR  Measures. 

12 

Inches    i: 

aakc 

1  Foot. 

3 

Feet 

u 

1  Yard. 

bh 

Yards 

u 

1  Rod  or  Pole. 

40 

Rods 

u 

1  Furlong. 

8 

Furlongs 

(i 

1  Mile. 

Also, 

7y^^^  Inches 

££ 

1  Link. 

25 

Links 

u 

1  Rod  or  Pole. 

100 

Luiks 

u 

1  Chain. 

10 

Chains 

u 

1  Furlong. 

8 

Furlongs 

(( 

1  Mile. 

Note.  —  For  other  linear  measures, 

see  National   Arithmetic,  Art. 

133,  134,  136. 

22 

254 


ELEMENTS   OF   GEOMETRY. 


611.     Table  of  Surface  Measures. 
144     Square  Inches  make  1  Square  Foot. 


9 

Square  Feet 

a 

1  Square  Yard. 

30|. 

Square  Yards 

a 

1  Square  Rod  or  Pole 

40 

Square  Rods 

u 

1  Rood. 

4 

Roods 

li 

1  Acre. 

640 

Acres 

a 

1  Square  Mile. 

625 

Square  Links 

u 

1  Square  Rod. 

16 

Square  Rods 

u 

1  Square  Chain. 

10 

Square  Chains 

a 

1  Acre. 

Also, 


612.  Since  an  acre  is  equal  to  10  chains,  or  100,000 
links,  square  chains  may  be  readily  reduced  to  acres  by 
pointingoff  one  decimal  place  from  the  right,  and  square 
links  by  pointing  off  five  decimal  places  from  the  right. 

Problem  I. 

613.  To  find  the  area  of  a  parallelogram. 

Multiply  the  base  by  the  altitude^  and  the  product  ivill 
be  the  area  (Prop.  Y.  Bk.  lY.). 

Examples.  p  q 

1.  What  is  the  area  of  a  square,  A  B  C  D, 
whose  side  is  25  feet  ? 

25  X  25  =  625  feet,  Ans. 

2.  What  is  the  area  of  a  square  field  whose 
side  is  35.25  chains  ?    Ans.  124  A.  1 R.  1  P. 

3.  How  many  square  feet  of  boards  are  required  to  lay 
a  floor  21  ft.  6  in.  square  ? 

4.  Required  the  area  of  a  square  farm,  whose  side  is 
3,525  links. 

5.  What  is  the  area  of  the  rectangle 
ABCD,  whose  length,  AB,  is  bQ  feet, 
and  whose  width,  AD,  is  37  feet  ? 

56  X  37  =  2,072  feet,  Ans. 


V> 


D 


B 


BOOK  XI.  255 

6.  How  many  square  feet  in  a  plank,  of  a  rectangular 
form,  which  is  18  feet  long  and  1  foot  6  inches  wide  ? 

7.  How  many  acres  in  a  rectangular  garden,  whose 
sides  are  326  and  153  feet  ?  Ans.  1  A.  23  P.  6^  yd. 

8.  A  rectangular  court  68  ft.  3  in.  long,  by  56  ft.  8  in. 
broad,  is  to  be  paved  with  stones  of  a  rectangular  form, 
each  2  ft.  3  in.  by  10  in.  ;  how  many  stones  will  be  re- 
quired ?  Ans.  2,062|  stones. 

9.  Required  the  area  of  the  rhom- 
boid A  B  C  D,  of  which  the  side  A  B 
is  354  feet,  and  the  perpendicular  dis- 
tance, E  F,  between  A  B  and  the  oppo- 
site side  C  D,  is  192  feet. 

354  X  192  =  67,968  feet,  Ans. 

10.  How  many  square  feet  in  a  flower-plat,  in  the  form 
of  a  rhombus,  whose  side  is  12  feet,  and  the  perpendicular 
distance  between  two  opposite  sides  of  which  is  8  feet  ? 

11.  How  many  acres  in  a  rhomboidal  field,  of  wliich  the 
sides  are  1,234  and  762  links,  and  the  perpendicular  dis- 
tance between  the  longer  sides  of  which  is  658  links  ? 

Ans.  8  A.  19  P.  4  yd.  6^  ft. 

Problem  II. 

614.  The  area  of  a  square  being  given,  to  find  the  side. 
Extract  the  square  root  of  the  area. 
Scholium.   This  and  the  two  following  problems  are  the 
converse  of  Prob.  I. 

Examples. 

1.  What  is  the  side  of  a  square  containing  625  square 
feet  ? 

a/ 625  =  25  feet,  the  side  required. 

2.  The  area  of  a  square  farm  is  124  A.  1  R.  1  P. ;  how 
many  links  in  length  is  its  side  ? 

3.  A  certain  corn-field  in  the  form  of  a  square  contains 


256  ELEMENTS  OF  GEOMETRY. 

15  A.  2  R.  20  p.  If  the  corn  is  planted  on  the  margin, 
4  hills  to  a  rod  in  lengtli,  how  many  hills  are  there  on 
the  margin  of  the  field  ?  Ans.  800  hills. 

Problem  III. 

615.  The  area  of  a  rectangle  and  either  of  its  sides 
being  given,  to  find  the  other  side. 

Divide  the  area  by  the  given  side,  and  the  quotient  will 

be  the  other  side. 

Examples. 

1.  The  area  of  a  rectangle  is  2,072  feet,  and  the  length 
of  one  of  the  sides  is  6Q  feet ;  what  is  the  length  of  the 
other  side  ? 

2072  H-  56  =  37  feet,  the  side  required. 

2.  How  long  must  a  rectangular  board  be,  which  is  15 
inches  in  width,  to  contain  11  square  feet  ? 

3.  A  rectangular  piece  of  land  containing  6  acres  is  120 
rods  long ;  what  is  its  width  ?  Ans.  8  rods. 

4.  The  area  of  a  rectangular  farm  is  266  A.  3  R.  8  P., 
and  the  breadth  46  chains  ;  what  is  the  length  ? 

Ans.  58  chains. 

Problem  IY. 

616.  The  area  of  a  rhomboid  or  rhombus  and  the  length 
of  the  base  being  given,  to  find  the  altitude  ;  or  the  area 
and  the  altitude  being  given,  to  find  the  base. 

Divide  the  area  by  the  length  of  the  base,  and  the  quo- 
tient will  be  the  altitude  ;  or  divide  the  area  by  the  alti- 
tude, and  the  quotient  will  be  the  length  of  the  base. 

Examples. 

1.  The  area  of  a  rhomboid  is  67,968  square  feet,  and 
the  length  of  the  side  taken  as  its  base  354  feet ;  what  is 
the  altitude  ? 

67,968  -^  354  =  192  feet,  the  altitude  required. 

2.  The  area  of  a  piece  of  land  in  the  form  of  a  rhombus 


BOOK  XI.  257 

is  69,452  square  feet,  and  the  perpendicular  distance  be- 
tween two  of  its  opposite  sides  is  194  feet ;  required  the 
length  of  one  of  the  equal  sides.  Ans.  358  ft. 

3.  On  a  base  12  feet  in  length  it  is  required  to  find  the 
altitude  of  a  rhomboid  containing  968  square  feet. 

4.  The  area  of  a  rhomboidal-shaped  park  is  lA.  3R. 
34  P.  5^  yd. ;  and  the  perpendicular  distance  between  the 
two  shorter  sides  is  96  yards ;  required  the  length  of  each 
of  these  sides  ?  Ans.  18  rods. 

Problem  V. 

617.  The  diagonal  of  a  square  being  given,  to  find  the 
area. 

Divide  the  square  of  the  diagonal  by  2,  and  the  quotient 
will  he  the  area,     (Prop.  XI.  Cor.  4,  Bk.  IV.) 

Examples.  D  C 

1.  The   diagonal,   AC,   of  the   square 
A  B  C  D,  is  30  feet ;  what  is  the  area  ? 

80^  =  900  ;  900  -^  2  =  450  square  feet, 
[the  area  required. 

B 

2.  The  diagonal  of  a  square  field  is  45  chains ;  how 
many  acres  does  it  contain  ? 

3.  The  distance  across  a  public  square  diagonally  is  27 
rods  ;  what  is  the  area  of  the  square  ? 

Problem  YI. 

618.  The  area  of  a  square  being  given,  to  find  the 
diagonal. 

Extract  the  square  root  of  double  the  area. 
Scholium.   This  problem  is  the  converse  of  the  last. 

Examples. 
1.  The  area  of  a  square  is  450  square  feet ;  what  is  its 
diagonal  ? 

450  X  2  =  900 ;  /v/"900  =  30  feet,  the  diagonal  required. 

22* 


258  ELEMENTS    OF   GEOMETRY. 

2.  The  area  of  a  public  square  is  4  A.  2  R.  9  P. ;  what 
is  the  distance  across  it  diagonally  ? 

3.  The  area  of  a  square  farm  is  57.8  acres  ;  what  is  the 
diagonal  in  chains  ?  Ans.  34  chains. 

Problem  YII. 

619.  The  sides  of  a  rectanCxLe  being  given,  to  cut  off  a 
given  area  by  a  line  parallel  to  either  side. 

Divide  the  given  area  by  the  side  which  is  to  retain  its 
length  or  width,  and  the  quotient  will  be  the  length  or 
width  of  the  part  to  be  cut  off.    (Prop.  lY.  Sch.,  Bk.  IV.) 

Examples. 

1.  If  the  sides  of  a  rectangle,  ABCD, 
are  25  and  14  feet,  how  wide  an  area, 
E  B  C  F,  to  contain  154  square  feet, 
can  be  cut  off  by  a  line  parallel  to  the 
bide  AD? 

154  -r- 14  =  11  feet,  the  width  required. 

2.  A  farmer  has  a  field  16  rods  square,  and  wishes  to 
cut  off  from  one  side  a  rectangular  lot  containing  exactly 
one  acre  ;  what  must  be  the  width  of  the  lot  ? 

3.  A  carpenter  sawed  off,  from  the  end  of  a  rectangular 
plank,  in  a  line  parallel  to  its  width,  5  square  feet.  From 
the  remainder  he  then  sawed  off',  in  a  line  parallel  to  the 
length,  8  square  feet.  Required  the  dimensions  of  the 
part  still  remaining,  provided  the  original  dimensions  of 
the  plank  were  20  feet  by  15  inches. 

Ans.  16  feet  by  9  inches. 

4.  The  length  of  a  certain  rectangular  lot  is  64  rods, 
and  its  width  50  rods  ;  how  far  from  the  longer  side  must 
a  parallel  line  be  drawn  to  cut  off  an  area  of  4  acres,  and 
how  far  from  the  shorter  side  of  the  remaining  portion  to 
cut  off  5  acres  and  2  roods  ?  How  many  acres  will  re- 
main after  the  two  portions  are  cut  off? 


BOOK  XI.  259 


Probleivi  YIII. 

620.  To  find  the  area  of  a  triangle,  the  base  and  alti- 
tnde  being  given. 

Multiply  the  base  by  half  the  altitude  (Prop.  \L 
Bk.  lY.). 

621.  Scholium.  The  same  result  can  be  obtained  by 
nuiltiplymg  the  altitude  by  half  the  base,  or  by  multiply- 
ing together  the  base  and  altitude  and  taking  half  the 
product. 

Examples. 

1.  Required  the  area  of  the  triangle 
A  B  C,  whose  base,  B  C,  is  210,  and  alti- 
tude, AD,  is  190  feet. 


190 

210  X     2^  =  19,950  square  feet,  the 
[area  required. 

ij  JJ  L" 

2.  A  piece  of  land  is  in  tlie  form  of  a  right-angled  tri- 
angle, having  the  sides  about  the  right  angle,  the  one  254 
and  the  other  136  yards ;  required  the  area  in  acres. 

Ans.  3A.  2R.  10  P.  29j^yd. 

3.  Required  the  number  of  square  feet  in  a  triangular 
board  whose  base  is  27  inches  and  altitude  27  feet. 

4.  What  is  the  area  of  a  triangle  whose  base  is  15.75 
chains,  and  the  altitude  10.22  chains? 

5.  What  is  the  area  of  a  triangular  field  whose  base  is 
97  rods,  and  the  perpendicular  distance  from  the  base  to 
the  opposite  angle  40  rods  ?  Ans.  12  A.  20  P. 

Problem  IX. 

622.  To  find  the  area  of  a  triangle,  the  three  sides 
being  given. 

From  half  the  sum  of  the  three  sides  subtract  each 


2G0 


ELEMENTS    OF   GEOMETRY, 


side ;  muUiply  the  half  sum  and  the  three  remainders  to- 
gether, and  the  square  root  of  the  product  will  be  the  area 
required. 

For, let  ABC  be  a  triangle  whose  three  c 

sides,  A B,  B  C,  AC,  are  given,  but  not 
the  altitude  C  D,  and  let  the  side  B  C  be 
represented  by  <z,  AC  by  6,  and  AB  by  c. 

Now,  since  A  is  an  acute  angle  of  the 
triangle  ABC,  we  have  (Prop.  XII. 
Bk.  IV.), 

fl^2  =  ^^2_|_  c'_2  6- X  AD,     or    AD  = 

Hence,  in  the  right-angled  triangle  ADC,  we  have  (Prop. 
XI.  Cor.  1,  Bk.  lY.), 


C  D^  =  ^2 


(&2  _|-  c2  _  «9)2  4  &2  c2  _  (62  ^ 


2\2 


«2) 


4c2 


4c9 


and,  by  extracting  the  square  root 
CD  =  ^^iZZ 


{b-^^r 


cfiyi 


2c 


But  the  area  of  the  triangle  ABC  is  equivalent  to  the 
product  of  c  by  half  of  C  D  (Prob.  YIII.)  ;  hence 


A  B  C  =  i  a/  4  tn-'-i  —  (69  -f-  ca  —  a2)a. 


The   expression  4  h'^ 


(h^  +  c2  —  ay,  being  the 


difference  of  two  squares,  can  be  decomposed  into 

(2  &  c  +  Z>^  +  c^  —  «0  X  (2bc  —  h^  —  6-  +  aO- 

Now,  the  first  of  these  factors  may  be  transformed  to 
(b  -\-  cy  —  a^,  and  consequently  may  be  resolved  into 
(^b  -\-  c  -\-  a')  X  (6  +  c  —  a)  ;  and  the  second  is  tlie 
same  thing  as  tt^ —  (ft  —  cy,  which  is  equal  to  (^a-\-b  —  c) 
X  (^a  —  6  +  6').     We  have  then 

^b^c^—(b''  +  c^  —  ay=  (a  +  b  +  c)  X  (b-\-c  —  a^ 
X  (a  +  c  —  b)  X  (a  +  b  —  c^. 


BOOK  XI.  261 

Let  S  represent  half  the  sum  of  the  three  sides  of  the 
triangle  ;  tlicu 

a-{.b+  c  =  2S;      b  +  c  —  a  =^2  (^  — a)  ; 
a+c  —  b  =  2(8  —  b);       a  +  b  —  c  =  2(S  —  c^; 
hence 

A  B  C  =  i  V  16  S  (S  —  «)  X  (S^^6)"xTS^=^)» 
which,  being  reduced,  gives  as  the  area  of  the  triangle,  as 
given  above, 

V  S(S  — a)  X  (S  — 6)  X  (S  — c). 

Examples.  C 

1.  What  is  the  area  of  a  triangle,  ABC, 
whose  sides,  AB,  BC,  CA,  are  40,  30, 
and  50  feet? 


30  +  40  +  50  4-  2    =  60,  half  the  sum  of  the  three  sides. 

60  —  30  =  30,  first  remainder. 

60  —  40  =  20,  second  remainder. 

60  —  50  =  10,  third  remainder. 
60  X  30  X   20   X  10  =  180,000;  Vl80~000  =  424.26 
square  feet,  the  area  required. 

2.  How  many  square  feet  in  a  triangular  floor,  whose 
sides  are  15,  16,  and  21  feet  ? 

3.  Required  the  area  of  a  triangular  field  whose  sides 
are  834,  658,  and  423  links. 

Ans.  lA.  IR.  20  P.  4  yd.  1.6  ft. 

4.  Required  the  area  of  an  equilateral  triangle,  of  which 
each  side  is  15  yards. 

5.  What  is  the  area  of  a  garden  in  the  form  of  a  paral- 
lelogram, whose  sides  are  432  and  263  feet,  and  a  diagonal 
342  feet  ?  Ans.  2  A.  10  P.  11.46  yd. 

6.  Required  the  area  of  an  isosceles  triangle,  whoso 
base  is  25  and  each  of  its  equal  sides  40  rods. 

7.  Wliat  is  the  area  of  a  rhoml)oidal  field,  whose  sides 
are  57  and  83  rods,  and  the  diagonal  127  lods  ? 

Ans.  22  A.  3  R.  21  P.  26  yd.  5  ft. 


262  ELEMENTS    OF   GEOMETRY. 

Problem  X. 

623.  Any  two  sides  of  a  right-angled  triangle  being 
given,  to  find  the  third  side. 

To  the  square  of  the  base  add  the  square  of  the  perpen- 
dicular ;  and  the  square  root  of  the  sum  will  give  the  hy- 
pothenvse  (Prop.  XI.  Bk.  lY.). 

From  the  square  of  the  hypothenuse  subtract  the  square 
of  the  given  side,  and  the  square  root  of  the  difference  loiil 
be  the  side  required  (Prop.  XI.  Cor.  1,  Bk.  IV.). 

Examples. 

1.  The  base,  AB,  of  the  triangle  ABC 
is  48  feet,  and  the  perpendicular,  B  C,  36 
feet ;  what  is  the  hypothenuse  ? 

48^  +  36^  =  3600  ;  V  3600  =  60  feet, 

[the  hypothenuse  required.      . 

2.  The  hypothenuse  of  a  triangle  is  53  feet,  and  the  per- 
pendicular 28  feet ;  what  is  the  base  ? 

3.  Two  ships  sail  from  tlie  same  port,  one  due  west  50 
miles,  and  the  other  due  south  120  miles ;  how  far  are 
they  apart  ?  Ans.  130  miles. 

4.  A  rectangular  common  is  25  rods  long  and  20  rods 
wide  ;  what  is  the  distance  across  it  diagonally  ? 

5.  If  a  house  is  40  feet  long  and  25  feet  wide,  with  a 
pyramidal-shaped  roof  10  feet  in  lieight,  how  long  is  a 
rafter  which  reaches  from  the  vertex  of  the  roof  to  a  cor- 
ner of  the  building  ? 

6.  There  is  a  park  in  the  form  of  a  square  containing 
1 0  acres  ;  how  many  rods  less  is  the  distance  from  the 
centre  to  each  corner,  than  the  length  of  the  side  of  the 
square  ?  Ans.  11.716  rods. 

Problem  XI. 
624.  The  sum  of  the  hypothenuse  and  perpendicular 


BOOK  XI.  263 

and  the  base  of  a  right-angled  triangle  being  given,  to 
find  the  hypothenuse  and  the  perpendicular. 

To  the  square  of  the  svm  add  the  square  of  the  base, 
and  divide  the  amount  by  twice  the  sum  of  the  hypothe- 
nuse  and  perpendicular ,  and  the  quotient  will  be  the  hy- 
pothenuse. 

From  the  sum  of  the  hypothenuse  and  perpendicidar 
subtract  the  hypothenuse,  and  the  remainder  will  be  the 
perpendicular. 

625.  Scholium.  This  problem  may  be  regarded  as  equiv- 
alent to  the  sum  of  two  numbers  and  the  difference  of 
their  squares  being  given,  to  find  the  numbers  (National 
Arithmetic,  Art.  553). 

Note.  —  The  learner  should  be  required  to  give  a  geometrical  demon- 
stration of  the  problem,  as  an  exercise  in  the  application  of  principles. 

Examples. 

1.  The  sum  of  the  hypothenuse  and  the  perpendicular 
of  a  right-angled  triangle  is  160  feet,  and  the  base  80  feet ; 
required  the  hypothenuse  and  the  perpendicular. 

Ans.  Hypothenuse,  100  ft. ;  perpendicular,  60  ft. 

1602  +  80^  =  32,000  ;  32,000  -f-  (160  X  2)  =  100 ; 
160  —  100  =  60. 

2.  Two  ships  leave  the  same  anchorage  ;  the  one,  sailing 
due  north,  enters  a  port  50  miles  from  the  place  of  depart- 
ure, and  the  other,  sailing  due  east,  also  enters  a  port,  but 
by  sailing  thence  in  a  direct  course  enters  the  port  of  the 
first ;  now,  allowing  that  the  second  passed  over,  in  all, 
90  miles,  how  far  apart  are  the  two  ports  ? 

3.  A  tree  100  feet  high,  standing  perpendicularly  on  a 
horizontal  plane,  was  broken  by  the  wind,  so  that,  as  it 
fell,  while  the  part  broken  off  remained  in  contact  with 
the  upright  portion,  the  top  reached  the  ground  40  feet 
from  tlie  foot  of  the  tree;  what  is  the  length  of  each  part? 

Ans.  The  part  broken  off,  58  ft. ;  the  upriglit,  42  ft. 


264  ELEMENTS  OF  GEOMETRY. 

Problem  XII. 

626.  The  area  and  the  base  of  a  triangle  being  given, 
to  find  the  altitude  ;  or  the  area  and  altitude  being  given, 
to  find  the  base. 

Divide  double  the  area  by  the  base,  and  the  qnotient 
will  be  the  altitude  ;  or  divide  double  the  area  by  the  alti- 
tude^ and  the  quotient  ivill  be  the  base. 

627.  Scholium.  This  problem  is  the  converse  of  Prob. 
VIII. 

Examples. 

1.  The  area  of  a  triangle  is  1300  square  feet,  and  the 
base  Qb  feet ;  what  is  the  altitude  ? 

1300  X  2  =  2600  ;  2600  -^  65  =  40  ft.,  altitude  required. 

2.  The  area  of  a  right-angled  triangle  is  17,272  yards,  of 
which  one  of  the  sides  about  the  right  angle  is  136  yards ; 
required  the  other  perpendicular  side. 

3.  The  area  of  a  triangle  is  46.25  chains,  and  the  alti- 
tude 5.2  chains  ;  what  is  the  base  ? 

4.  A  triangular  field  contains  30  A.  3  R.  27  P.  ;  one  of 
its  sides  is  97  rods  ;  required  the  perpendicular  distance 
from  the  opposite  angle  to  that  side.  Ans.  102  rods. 

Problem  XIII. 

628.  To  find  the  area  of  a  trapezoid. 

Multiply  half  the  sum  of  its  parallel  sides  by  its  altitude 
(Prop.  YII.  Bk.  lY.). 

Examples.  DEC 

1.  What  is  the  area  of  the  trapezoid  / 

ABCD,   whose   parallel  sides,  A  B,         / 

D  C,  are  32  and  24  feet,  and  the  alti-      [_ 

tude,  E  F,  20  feet  ?  A  F  B 

32  +  24  =  bQ)  56  ~  2  =  28 ;  28  X  ^0  =  560  ^q.  ft., 

[tlie  area  required. 


BOOK  XI.  205 

2.  How  many  square  feet  in  a  board  in  the  form  of  a 
trapezoid,  whose  width  at  one  end  is  2  feet  3  inches,  and 
at  tlie  other  1  foot  6  inches,  the  length  being  16  feet  ? 

3.  Required  the  area  of  a  garden  in  the  form  of  a  trape- 
zoid, wliose  parallel  sides  are  786  and  473  links,  and  the 
perpendicular  distance  between  them  986  links. 

Ans.  6A.  38P.  3  yd. 

4.  How  many  acres  in  a  quadrilateral  field,  having  two 
parallel  sides  83  and  101  rods  in  length,  and  which  are 
distant  from  each  other  60  rods  ? 

Problem  XIV. 

629.  To  find  the  area  of  a  regular  polygon,  the  pe- 
rimeter and  apothegm  being  given. 

Multiply  the  perimeter  by  half  the  apothegm^  and  t/ie 
product  will  be  the  area  (Prop.  VIII.  Bk.  VI.). 

630.  Scholium.  This  is  in  effect  resolving  the  polygon 
into  as  many  equal  triangles  as  it  has  sides,  by  drawing 
lines  from  the  centre  to  all  the  angles,  then  finding  their 
areas,  and  taking  their  sum. 

Examples, 

1.  Bequired  the  area  of  a  regu- 
lar hexagon,  A  B  C  D  E  F,  whose 
sides,  A  B,  B  C,  &c.  are  each  15 
yards,  and  the  apothegm,  O  M,  13 
yards. 

15X6  =  90;  90-^V-  =  585yd.,  ____ 

[the  area  required.  A     M     B 

2.  What  is  the  area  of  a  regular  pentagon,  whose  sides 
are  each  25  feet,  and  the  perpendicular  from  the  centre  to 
a  side  17.205  feet  ? 

3.  A  park  is  laid  out  in  the  form  of  a  regular  heptagon, 
whose  sides  are  each  19.263 chains  ;  and  the  perpendicular 

23 


2G6 


ELEMENTS    OF   GEOMETRY. 


distance  from  the  centre  to  each  of  the  sides  is  20  chains. 
How  many  acres  does  it  contain  ? 

Ans.  134  A.  311.  14  P. 


Problem  XY. 

631.  To  find  the  area  of  a  regular  polygon,  its  side  or 
perimeter  being  given. 

Multiply  the  square  of  the  side  of  the  polygon  hy  the 
area  of  a  similar  polygon  whose  side  is  unity  or  1  (Prop. 
XXXI.  Bk.  IV.). 

632.  A  Table  of  Regular  Polygons  whose  Side  is  1. 


NAMES. 

AREAS. 

NAMES. 

AUEAS. 

Triangle, 

Square, 

Pentagon, 

Hexagon, 

Heptagon, 

0.4330127 
1.0000000 
1.7204774 
2.5980762 
3.6339124 

Octagon, 

Nonagon, 

Decagon, 

Undecagon, 

Dodecagon, 

4.8284271 
6.1818242 
7.6942088 
9.3656399 
11.1961524 

The  apothegm  of  any  regular  polygon  wliose  side  is 
1  being  ascertained,  its  area  is  computed  readily,  by 
Prob.  XIY. 

Examples. 

1.  Required  the  area  of  an  equilateral  triangle,  whose 
side  is  100  feet. 

100=  =  10,000  ;  10,000  X  0.4830127  =  4380.127  square 

[feet,  the  area  required. 

2.  What  is  the  area  of  a  regular  pentagon,  whose  side 
is  87  yards  ? 

8.  How  many  acres  in  a  field  in  the  form  of  a  regular 
undecagon,  whose  side  is  27  yards  ? 

Ans.  1  A.  1 R.  25  P.  21  yd.  2.7  ft. 


BOOK  XI.  267 

4.  What  is  the  area  of  an  octagonal  floor,  whose  side  is 
15  ft.  6  in.  ? 

5.  How  many  acres  in  a  regular  nonagon,  whose  perim- 
eter is  2286  feet  ?  Ans.  9  A.  24  P.  28  yd. 

Problem   XYI. 

633.  To  find  the  side  of  any  regular  polygon,  its  area 
heing  given. 

Divide  the  given  area  by  the  area  of  a  similar  polygon 
whose  side  is  1,  and  the  square  root  of  the  quotient  ivill 
be  the  side  required. 

634.  Scholium,  This  problem  is  the  converse  of  Prob. 
XY. 

Examples. 

1.  The  area  of  an  equilateral  triangle  is  4330.127  square' 
feet ;  what  is  its  side  ? 

4330.127  -^  .4330127  =  10,000;  V  10,000  =  100  feet, 

[the  side  required. 

2.  The  area  of  a  regular  hexagon  is  1039.23  feet ;  what 
is  its  side  ? 

3.  The  area  of  a  regular  decagon  is  7  P.  18  yd.  5  ft. 
128.55  in. ;  what  is  its  side  ?  Ans.  16  ft.  5  in. 

Problem  XVII. 

635.  To  find  the  area  of  an  irregular  polygon. 
Divide   the  polygon  into  triangles,   or   triangles  and 

trapezoids,  and  find  the  areas  of  each  of  them  separately  ; 
the  sum  of  these  areas  will  be  the  area  required. 

636.  Scholium.  When  the  irregular  polygon  is  a  quad- 
rilateral, the  area  may  be  found  by  multiplying  together 
the  diagonal  and  lialf  the  sum  of  the  perpendiculars  drawn 
from  it  to  the  opposite  angles. 


268  ELEMENTS   OF   GEOMETRY. 

Examples. 

1.  Kequired  the  area  of  the  irregular 
pentagon  A  B  C  D  E,  of  which  the  diag- 
onal AC  is  20  feet,  and  A D  30  feet ; 
and  the  perpendicular  distance  from 
the  angle  B  to  A  C  is  8  feet,  from  C  to 
AD  12  feet,  and  from  E  to  AD  6  feet. 

20  X  I  =  80  ;      36  X  ¥-  =  216 ;      36  X  |  =  108 ; 
80  +  216  +  108  =  504  sq.  ft.,  the  area  required. 

2.  What  is  the  area  of  a  trapezium,  whose  diagonal  is 
42  feet,  and  the  two  perpendiculars  from  the  diagonal  to 
the  opposite  angles  are  16  and  18  feet  ? 

3.  In  an  irregular  hexagon,  A  B  C  D  E  F,  are  given  the 
sides  A  B  536,  B  C  498,  CD  620,  DE  580,  EF  308,  and 
AF  492  linlfs,  and  the  diagonals  AC  918,  CE  1048,  and 
A  E  652  links  ;  required  the  area. 

Ans.  6A.  2R.  9P.  23  yd.  8.4  ft; 

4.  In  measuring  along  one  side,  AB,  of  a  quadrangular 
field,  A  B  C  D,  that  side  and  the  perpendiculars  let  fall  on 
it  from  two  opposite  corners  measured  as  follows :  A  B 
1110,  AE  110,  AF  745,  DE  352,  CF  595  links.  What 
is  the  area  of  the  field  ?         Ans.  4  A.  1 R.  5  P.  24  yd. 

5.  In  a  four-sided  rectilineal  field,  A  B  C  D,  on  account 
of  obstructions,  there  could  be  taken  only  the  following 
measures  :  the  two  sides  B  C  265  and  AD  220  yards,  the 
diagonal  A  C  378,  and  the  two  distances  of  the  perpendic- 
ulars from  the  ends  of  the  diagonal,  namely,  A  E  100,  and 
C  F  70  yards.     Required  the  area  in  acres. 

Problem  XVIII. 

637.  To  find  the  circumference  of  a  circle,  wlien  the 
diameter  is  given,  or  the  diameter  when  the  circumference 
is  given. 

Multiply  the  diameter  hy  3.1416,  and  the  product  vnll 
he   the    circumference ;   or,  divide  the  circumference  by 


BOOK  XT.  269 

3.1416,  and  the  quotient  tvill  he  the  diameter  (Prop.  XY. 
Cor.  3,  Bk.  YI.). 

638.  Scholium,  The  diameter  may  also  be  found  by 
multiplying  the  circumference  by  .31831,  the  reciprocal 
of  3.1416. 

Examples. 

1.  The  diameter,  AB,  of  the  cir- 
cle A  E  B  F  is  100  feet ;  what  is 
its  circumference  ? 

100  X  3.1416  ==  314.16  feet,  the 
[circumference  required. 

2.  Required  the  circumference  of  a  circle  whose  diam- 
eter is  628  luiks.  Ans.  Ifur.  38  rd.  5  yd.  1.56  in. 

3.  If  the  diameter  of  the  earth  is  7912  miles,  what 
is  its  circumference  ? 

4.  Required  the  diameter  of  a  circular  pond  whose  cir- 
cumference is  928  rods. 

Ans.  7  fur.  15  rd.  2  yd.  5.55  in. 

5.  The  circumference  of  a  circular  garden  is  1043  feet ; 
what  is  its  radius  ?  Ans.  10  rd.  1  ft. 

Problem  XIX. 

639.  To  find  the  length  of  an  arc  of  a  circle  containing 
any  number  of  degrees,  the  radius  or  diameter  being  given. 

Multiply  the  number  of  degrees  in  the  given  arc  by 
0.01745,  and  the  product  by  the  radius  of  the  circle. 

For,  when  the  diameter  of  a  circle  is  1,  the  circumfer- 
ence is  3.1416  (Prop.  XY.  Sch.  1,  Bk.  YI.)  ;  hence,  when 
the  radius  is  1,  the  circumference  is  6.2832  ;  which,  divided 
by  360,  the  number  of  degrees  into  which  every  circle  is 
supposed  to  be  divided,  gives  0.01745,  the  length  of  the 
arc  of  1  degree,  wlien  the  radius  is  1. 

640.  Scholium.   Each  of  the  360  degrees  of  a  circle, 

23* 


270  ELEMENTS  OP  GEOMETRY. 

marked  tlius,  360°,  is  divided  into  60  minutes,  marked 
thus,  60',  and  each  minute  into  60  seconds,  marked  thus, 
60"  (National  Arithmetic,  Art.  1-18). 

Examples. 

1.  What  is  the  length  of  an  arc, 
A  D,   containing  60°  30'  on  the  cir- 
cumference of  a  circle  whose  radius, 
AC,  is  100  feet? 
60°  30'  =  60.5° ;  60.5  X  0.01745  = 

1.055725;    1.055725  x  100   = 
105.5725  ft.,  arc  required. 

2.  Required  the  length  of  an  arc  of  31°  15',  the  radius 
being  12  yards. 

3.  Required  tlie  lengtli  of  an  arc  of  12°  10',  the  diam- 
eter being  20  feet.  Ans.  2.1231  feet. 

4.  What  is  the  length  of  an  arc  of  57°  17'  44i",  the 
radius  being  25  feet  ?  Ans.  25  feet. 

Problem  XX. 

641.  To  find  the  area  of  a  circle. 

Multiply  the  circumference  by  half  the  radius  (Prop. 
XV.  Bk.  VI.)  ;  or,  multiply  the  square  of  the  radius  by 
3.1416  (Prop.  XV.  Cor.  2,  Bk  VI.). 

642.  Scholium.  Multiplying  the  circumference  by  half 
the  radius  is  the  same  as  multiplying  the  circumference 
and  diameter  together,  and  taking  one  fourth  of  the  pro- 
duct. Now,  denoting  the  circumference  by  c,  and  the 
diameter  by  d,  since  c  =  3.1416  X  d  (Prob.  XVIII.),  we 
have  (d  X  3.1416  x  d) -^  4  =  d^  X  0.7854  =  the  area 
of  a  circle.  Again,  since  d=c-T- 3.1416  (Prob.  XVIII.), 
we  have  c  ^  3.1416  X  c  -^  4  =  c^  -^  12.5664,  which  is, 
by  taking  the  reciprocal  of  12.5664,  equal  to  c^  X  0.07958 
=  the  area  of  the  circle.  Hence  the  area  of  the  circle 
may  also  be  found  by  multiplying'  the  square  of  the  diam- 


BOOK    XI.  271 

eler  by  0.7854 ;  or  by  multiplying^  the  square  of  the  cir- 
cumference by  0.07958. 

Examples. 

1.  The  circumference  of  a  circle  is  814.16  feet,  and  its 
radius  50  feet ;  what  is  its  area  ? 

814.16  X  ^i-  =  7854  feet,  the  area  required. 

2.  If  the  circumference  of  a  circle  is  855  feet,  and  its 
diameter  113  feet,  what  is  the  area? 

8.  What  is  the  area  of  a  circular  garden,  whose  radius 
is  281 J  links  ?  Ans.  2  A.  1 R.  88  rd.  9yd.  5  ft. 

4.  A  horse  is  tethered  in  a  meadow  by  a  cord  89.25075 
yards  long  ;  over  how  much  ground  can  he  graze  ? 

5.  Required  the  area  of  a  semicircle,  the  diameter  of 
the  whole  circle  beinji  751  feet. 

Ans.  5  A.  18  P.  16  yd. 


•«> 


Problem  XXI. 

643.  To  find  the  diameter  or  circumference,  the  area 
being  given. 

Divide  the  area  by  0.7854,  and  the  square  root  of  the 
quotient  will  be  the  diameter;  or^  divide  the  area  by 
0.07958,  and  the  square  root  of  the  quotient  will  he  the 
circumference . 

644.  Scholium,  This  problem  is  the  converse  of  Prob. 
XX. 

EXAMPLES- 

1.  The  area  of  a  circle  is  814,16  feet ;  what  is  the 
diameter  ? 

814.16  ^  0.7854  =  400  ;  V  400  =  20  feet,  the  diameter 

[required. 

2.  What  must  be  the  length  of  a  cord  to  be  used  as  a 
radius  in  describing  a  circle  which  shall  contain  exactly 
1  acre  ? 

8.  The  area  of  a  circular  pond  is  6  A.  1 R.  27  P. 
18.2  yd.  ;  what  is  the  circumference  ?  Ans.  625  yd. 


272  ELEMENTS  OF  GEOMETRY. 

4.  The  area  of  a  circle  is  T856  feet ;  what  is  the  cir- 
cumference ? 

5.  The  length  of  a  rectangular  garden  is  32,  and  its 
width  18  rods  ;  required  the  diameter  of  a  circular  garden 
having  the  same  area.  Ans.  27  rd.  1  ft.  4  in. 

Problem  XXII. 

645.  To  find  the  area  of  a  sector  of  a  circle. 
MuUiphj  the  arc  of  the  sector  by  half  of  its  radius 

(Prop.  XV.  Cor.  1,  Bk.  VI.)  ;  or, 

As  3G0''  are  to  the  degrees  in  the  arc  of  the  sector,  so  is 
the  area  of  the  circle  to  the  area  of  the  sector. 

Examples. 

1.  Required  the  area  of  a  sector, 
D  E,  whose  arc  is  80  feet,  and  its 
radius,  0  E,  70  feet. 

80  X  -V-  =  2800  square  feet,  the  area 

[required. 

2.  Required  the  area  of  a  sector,  of  which  the  arc  is  90 
and  the  radius  112  yards. 

3.  Required  the  area  of  a  sector,  of  which  the  angle  is 
137"  20',  and  the  radius  456  links. 

Ans.  2  A.  IR.  38  P.  21.92  yd. 

Problem  XXIII. 

646.  To  find  the  area  of  a  segment  of  a  circle. 

Find  the  area  of  the  sector  having-  the  same  arc  ivith 
the  segment,  and  also  the  area  of  the  triangle  formed  by 
the  chord  of  the  segment  and  the  radii  of  the  sector. 
Then,  if  the  segment  is  less  than  a  semicircle,  take  the 
difference  of  these  areas  ;  but  if  greater,  take  their  sum. 

647.  "Scholium.   When  the  heiglit  of  the  segment  and 


BOOK   XI. 


273 


the  diameter  of  the  circle  are  given,  the  area  may  be 
readily  found  by  means  of  a  table  of  segments,  by  divid- 
ing the  height  by  the  diameter^  and  looking  in  the  table 
for  the  quotient  in  the  column  of  heights,  and  taking  ovt, 
in  the  next  column  on  the  right  hand,  the  corresponding 
area;  lahich,  multiplied  by  the  square  of  the  diameter, 
will  give  the  area  required. 

When  the  quotient  cannot  be  exactly  found  in  the  table, 
proportions  may  be  instituted  so  as  to  find  the  area  be- 
tween the  next  higher  and  the  next  lower,  in  the  same 
ratio  that  the  given  height  varies  from  the  next  higher 


and  lower  heights. 


648.    Table  of  Segments. 


*s 

a 

.01 

Seg. 
Area. 

i  ^ 

;  tp 
'S 

03 

Seg. 
Area. 

33 

Seg. 
Area. 

[3) 

'S 

!I3 

.31 

Seg. 
Area. 

.41 

Seg. 
Area. 

.00133 

.11 

.04701 

.21 

.11990 

.20738 

.30319 

.02 

.00375 

.12 

.05339 

.22 

.12811 

.32 

.21667 

.42 

.31304 

.03 

.00687 

;.i3 

.06000 

.23 

.13646 

.33 

.22603 

.43 

.32293 

.04 

.01054 

.14 

.06683 

.24 

.14494 

.34 

.23547 

.44 

.33284 

.05 

.01468 

.15 

.07387 

.25 

.15354 

.35 

.24498 

.45 

.34278 

.06 

.01924 

.16 

.08111 

.26 

.16226 

.36 

.25455 

.46 

.35274 

.07 

.02417 

.17 

.08853 

.27 

.17109 

.37 

.26418 

.47 

.36272 

.08 

.02944 

.18 

.09613 

.28 

.18002 

.38  .27386 

.48 

.37270 

.09 

.03502 

.19 

.10390 

.29 

.18905 

.39  .28359 

.49 

.38270 

.10 

.04088 

.20 

.11182 

.30 

.19817 

.40  .29337 

1 

.50 

.39270 

^The  segments  in  the  table  are  those  of  a  circle  whoso 
diameter  is  1,  and  the  first  column  contains  the  corre- 
sponding heights  divided  by  the  diameter.  The  method 
of  calculating  the  areas  of  segments  from  the  elements  in 
the  table  depends  upon  the  principle  that  similar  plane 
figures  are  to  each  other  as  the  squares  of  their  like  linear 
dimensions. 

Examples. 
1.  Wliat  is   the  area  of  the  segment  ABE,   its   arc 
A  E  B  being  73.74%   its   chord  A  B  being   12  feet,   and 


274  ELEMENTS  OF  GEOMETRY. 

the   radius,    C  B,    of    the    circle    10 
feet? 

0.7854  X  20'  =  314.16,  area  of  circle ; 
then  3G0° :  73.74° : :  314.16  :  64.3504, 
area  of  sector  A  E  B  C  ;  and,  by  Prob- 
lem IX.,  48  is  the  area  of  the  trian- 
gle ABC;  64.3504  —  48  =  16.3504 
feet,  the  area  required. 

2.  Eequired  the  area  of  a  segment  whose  height  is  18, 
and  the  diameter  of  the  circle  50  feet. 

18  -7-50  =  .36  ;  to  which  the  corresponding  area  in  the  ta- 
ble is  .25455  ;  .25455  X  50^  =  636.375,  area  required. 

3.  Required  the  area  of  a  segment  whose  arc  is  100°, 
chord  153.208  feet,  and  the  diameter  of  the  circle  200  feet. 

4.  What  is  the  area  of  a  segment  whose  height  is  4  feet, 
and  the  radius  51  feet  ?  Ans.  106  feet. 

5.  Required  the  area  of  a  segment,  the  arc  being  160°, 
chord  196.9616  feet,  and  the  radius  of  the  circle  100  feet. 

Problem  XXIY. 

649.  To  find  tlie  area  of  a  circular  zone,  or  tlie  space 
included  between  two  parallel  chords  and  their  intercepted 
arcs. 

From  the  area  of  tlie  inliole  circle  subtract  the  areas  of 
the  segments  on  the  sides  of  the  zone. 

Examples. 

1.  What  is  the  area  of  a  zone  whose  chords  are  each  12 
feet,  subtending  eacli  an  arc  of  73.74°,  when  the  radius  of 
the  circle  is  10  feet  ? 

Area  of  the  whole  circle  by  Prob.  XX.  =  314.16 ;  area 
of  each  segment  by  Prob.  XXIII.  =  16.3504  ;  16.3504 
X  2  =  32.7008  =  area  of  both  segments ;  314.16  — 
32.7008  =:  281.4592,  the  area  required. 


BOOK   XI.  i^K) 

2.  What  is  the  area  of  a  circular  zone  whose  longer 
chord  is  20  yards,  subtending  an  arc  of  60°,  and  the  shorter 
chord  14.66  yards,  subtending  an  arc  of  43°,  the  diameter 
of  tlie  circle  being  40  yards  ? 

3.  A  circle  whose  diameter  is  20  feet  is  divided  into 
three  parts  by  two  parallel  chords ;  one  of  the  segments 
cut  off  is  8  feet  in  height,  and  the  other  6  feet ;  what  is 
the  area  of  the  circular  zone  ?  Ans,  117.544  ft. 

Problem  XXV. 

650.  To  find  the  area  of  a  crescent. 

Find  the  difference  of  the  areas  of  the  two  segments 
formed  hy  the  arcs  of  the  crescent  and  its  chord. 

Examples. 

1.  The  arcs  A  C  B,  A  E  B,  of 
circles  having  the  same  radius, 
50  rods,  intersecting,  form  the 
crescent  A  C  B  E  ;  the  height, 
D  C,  of  the  segment  A  C  B  is  60 
rods,  and  the  height,  D  E,  of  the  segment  ABE  is  40 
rods  ;  what  is  the  area  of  the  crescent  ? 

The  area  of  the  segment  A  C  B,  by  Prob.  XXIII.,  is  4920.3 
rods,  and  that  of  the  segment  A  B  E  is  2933.7  rods  ; 
4920.3  —  2933.7  =  1986.6  rods,  the  area  of  the  crescent. 

2.  If  the  arc  of  a  circle  whose  diameter  is  24  yards  in- 
tersects a  circle  whose  diameter  is  20  yards,  forming  a 
crescent,  so  that  the  height  of  the  segment  of  the  first  cir- 
cle is  5.072  yards,  an-d  that  of  the  segment  of  the  second 
circle  is  8  yards,  what  is  the  area  of  the  crescent  ? 

Problem  XXVI. 

651.  To  find  the  area  of  a  circular  ring,  or  the  space 
included  between  two  concentric  circles. 

Find  the  areas  of  the  two  circles  separately  (Prob. 
XX.),  and  take  the  difference  of  these  areas;  or  sub- 


276  ELEMENTS   OF  GEOMETRY. 

tract  the  square  of  the  less  diameter  from  the  square  of 
the  g-reater^  and  multiply  their  difference  by  0.T851  (Prob. 
XX,  Sell.). 

Examples, 

1.  Required  the  area  of  the  ring  formed  by  two  circles 
■\yhose  diameters  are  30  and  50  feet. 

50-^— £0^  =  1400;  1400  X  0.7854=1099.56  sq.  feet, 

[tlie  area  of  the  ring. 

2.  What  is  tlie  area  of  a  ring  formed  by  two  circles 
whose  radii  are  36  and  24  feet  ? 

3.  A  circular  park,  256  yards  in  diameter,  has  a  car- 
riage-way running  around  it  29  feet  wide ;  what  is  tlie 
area  of  the  carriage-way  1 

Ans.  lA.  2R.  26P.  21.5  yd. 

Problem  XXVII. 

652.  The  diameter  or  circumference  of  a  circle  being 
given,  to  find  the  side  of  an  equivalent  square. 

Multiply  the  diameter  hy  0.8862,  or  the  circumference 
by  0.2821 ;  the  product  in  either  case  will  be  the  side  of 
an  equivalent  square. 

For,  since  0.7854  is  the  area  of  a  circle  whose  diameter 
is  1  (Prob.  XX.  Sch.),  the  square  root  of  0.7854,  which  is 
0.8862,  is  the  side  of  a  square  wliich  is  equivalent  to  a 
circle  whose  diameter  is  1.  Now  when  the  circumference 
is  1,  the  side  of  an  equivalent  square  must  have  the  same 
ratio  to  0.8862  as  the  diameter  1  has  to  its  circumference 
3.1416  (Prop.  XV.  Cor.  4,  Bk.  VI.)  ;  and  0.8862-^-3.1416 
gives  0.2821  as  the  side  of  the  equivalent  square  when 
the  circumference  is  1, 

ExAMPij:s. 
1.  The  diameter  of  a  circle  is  120  feet ;  what  is  the  side 
of  an  equivalent  square  ? 

120  X  0.8862  =  106.344  feet,  the  side  required. 


BOOK  XI.  277 

2.  The  circumference  of  a  circle  is  100  yards  ;  what  is 
the  side  of  an  equivalent  square  ?  Ans.  28.21yd. 

3.  There  is  a  circular  floor  30  feet  in  diameter ;  what 
is  the  side  of  a  square  floor  containing  the  same  area  ? 

4.  If  500  feet  is  the  circumference  of  a  circular  island, 
what  is  the  side  of  a  square  of  equal  area  ? 

Ans.  141.05  ft. 

Problem  XXVIII. 

653.  The  diameter  or  circumference  of  a  circle  being 
given,  to  find  the  side  of  the  inscribed  square. 

Multiplij  the  diameter  by  0.7071,  or  the  circumference 
by  0.2251 ;  the  product  in  either  case  will  be  the  side  of 
the  inscribed  square. 

For  0.7071  is  tlie  side  of  the  inscribed  square  when  the 
diameter  of  the  circumscribed  circle  is  1,  since  the  side  of 
the  inscribed  square  is  to  the  radius  of  the  circle  as  the 
square  root  of  2  to  1  (Prop.  IV.  Cor.,  Bk.  VI.)  ;  conse- 
quently, the  side  is  to  the  diameter,  or  twice  the  radius,  as 
half  the  square  root  of  2  is  to  1,  and  half  the  square  root 
of  2  is  0.7071,  approximately.  Now,  the  ratio  of  the 
diameter  of  a  circle  to  the  side  of  its  inscribed  square 
being  as  1  to  0.7071,  and  the  ratio  of  the  circumference  of 
a  circle  to  its  diameter  as  3.1416  to  1,  the  ratio  of  the 
inscribed  square  is  to  the  circumference  of  the  circle  as 
0.7071  to  3.1416  ;  and  0.7071  -^  3.1416  gives  0.2251  as 
the  side  of  the  inscribed  square  when  the  circumference 
is  1. 

Examples. 

1.  The  diameter,  AC,  of  a  circle  is 
110  feet ;  what  is  the  side,  A  B,  of  the 
inscribed  square  ? 

110  X  0.7071  =  77.781  feet,  the  side 

[required. 

24 


278  ELEMENTS  OF  GEOMETRY. 

2.  The  circumference  of  a  circle  is  300  feet ;  what  is 
the  side  of  the  inscribed  square  ?  Aiis.  67.53  ft. 

3.  A  log  is  36  inches  in  diameter  ;  of  how  many  inches 
square  can  a  stick  be  hewn  from  it  ? 

4.  There  is  a  circular  field  1000  rods  in  circuit ;  what 
is  the  side  of  the  largest  square  that  can  be  described  in 
it  ?  Ans.  225.10  rods. 

Problem  XXIX. 

654.  The  diameter  or  circumference  of  a  circle  beinir 
given,  to  find  the  side  of  an  inscribed  equilateral  tri- 
angle. 

MuUipli/  the  diameter  by  0.8660,  or  the  circvmference 
by  0.2757 ;  the  product  in  either  case  will  be  the  side  of 
the  inscribed  equilateral  triangle. 

For  0.8660  is  the  side  of  the  inscribed  equilateral  trian- 
gle when  the  diameter  of  the  circumscribed  circle  is  1, 
since  the  side  of  the  inscribed  equilateral  triangle  is  to  the 
radius  of  the  circle  as  the  square  root  of  3  is  to  1  (Prop. 
V.  Cor.  3,  Bk.  YI.)  ;  consequently,  the  side  is  to  the  diam- 
eter, or  twice  the  radius,  as  half  the  square  root  of  3  is  to 
1,  and  half  the  square  root  of  3  is  0.8660,  approximately. 
Also,  since  the  ratio  of  the  circumference  of  a  circle  to 
its  diameter  is  as  3.1416  to  1,  the  side  of  the  inscribed 
equilateral  triangle,  when  the  circumference  is  1,  equals 
0.8660  4-  3.1416,  or  0.2757. 

Examples. 

1.  Required  the  side  of  an  equilateral  triangle  that  may 
be  inscribed  in  a  circle  101  feet  in  diameter. 

101  X  0.8660  =  87.4660  feet,  tlie  side  required. 

2.  Required  the  side  of  an  equilateral  triangle  that  may 
be  inscribed  in  a  circle  80  rods  in  circumference. 

Ans.  22.05  rods, 

3.  Required  the  side  of  the  largest  equilateral  triangular 
beam  that  can  be  hewn  from  a  piece  of  round  timber  36 
inches  in  diameter. 


BOOK  XI.  279 

4.  Required  the  side  of  an  equilateral  triangle  that  can 
be  inscribed  in  a  circle  251.33  feet  in  circumference. 

5.  How  much  less  is  the  area  of  an  equilateral  triangle 
that  can  be  inscribed  in  a  circle  100  feet  in  diameter,  than 
the  area  of  the  circle  itself?  Ans.  4606.4  sq.  ft. 

The  Ellipse.  „ 

655.  An  Ellipse  is  a  plane  figure  bounded  by^curycr;"   ^^ 
from  any  point  of  which  the  sum  of  the  distances  to  two 
fixed  points  is  equal  to  a  straight  line  drawu  through 
those  two  points,  and  terminated  both  ways  by  the  curve. 

Thus  A  D  B  C  is  an  ellipse.  The 
two  fixed  points  G  and  H  are  called 
the  foci.  The  longest  diameter,  AB, 
of  the  ellipse  is  called  its  major  or 
transverse  axis,  and  its  sliortest  di- 
ameter, CD,  is  called  its  minor  ov 
conjugate  axis. 

QbQ.  The  AREA  of  an  ellipse  is  a  mean  proportional 
between  the  areas  of  two  circles  whose  diameters  are  the 
two  axes  of  the  ellipse. 

This,  however,  can  only  be  well  demonstrated  by  means 
of  Analytical  Geometry,  a  branch  of  tlie  matliematics  with 
which  the  learner  here  is  not  supposed  to  be  acquainted^ 

Problem  XXX. 

657.  To  find  the  area  of  an  ellipse,  the  major  and 
minor  axes  being  given. 

Multiply  the  axes  together,  and  their  product  by  0.7854, 
and  the  result  will  be  the  area. 

For  AB^  X  0.7854  expresses  the  area  of  a  circle  whose 
diameter  is  AB,  and  C  D^  X  0.7854  expresses  the  area  of 
a  circle  whose  diameter  is  C  D  ;  and  the  product  of  these 
two  areas  is  equal  to  A  B-*  X  C  D^  X  0.7854^,  which  is 


280  ELEMENTS  OP  GEOMETRY. 

equal  to  the  square  of  A  B  X  C  D  X  0.7854  ;  hence, 
A  B  X  C  D  X  0.7854  is  a  mean  proportional  between 
the  areas  of  the  two  circles  whose  diameters  are  A  B  and 
C  D  (Prop.  TV.  Bk.  II.)  ;  consequently  it  measures  the 
area  of  an  ellipse  whose  axes  are  AB  and  CD  (Art.  QdQ}. 

Examples. 

1.  Required  the  area  of  an  ellipse,  of  which  the  major 
axis  is  60  feet,  and  the  minor  axis  40  feet. 

60  X  40  X  0.7854  =  1884.96  sq.  ft.,  the  area  required. 

2.  "What  is  the  area  of  an  ellipse  whose  axes  are  75  and 
85  feet  ? 

8.  Required  the  area  of  an  ellipse  whose  axes  are  526 
and  854  inches.  Ans.  112  yd.  7  ft.  84.62  in. 

4.  How  many  acres  in  an  elliptical  pond  whose  semi- 
axes  are  436  and  254  feet  ? 

Ans.  7A.  8R.  37  P.  27  yd.  7  ft. 


BOOK   XII. 

APPLICATIONS   OF    GEOMETRY   TO  THE    IVIENSU- 
RATION   OF   SOLIDS. 

DEFINITIONS. 

658.  Mensuration  op  Solids,  or  Volumes,  is  the  pro- 
cess of  determining  their  contents. 

The  SUPERFICIAL  CONTENTS  of  a  body  is  its  quantity  of 
surface. 

Tlie  SOLID  CONTENTS  of  a  body  is  its  measured  magni- 
tude, volume,  or  solidity. 

659.  The  unit  of  volume,  or  solidity,  is  a  cube,  whose 
faces  are  each  a  superficial  unit  of  the  surface  of  the  body, 
and  whose  edges  are  each  a  linear  unit  of  its  linear  di- 


mensions. 

660. 

Table  of  1 

Solid  Measures. 

1728   Cubic 

Inches  make  1  Cubic  Foot 

27 

u 

Feet 

u 

1      "      Yard. 

4492i 

u 

Feet 

u 

1      "      Rod. 

32,768,000 

u 

Rods 

a 

1      "      Mile. 

Also, 

231 

u 

Inches 

a 

1  Liquid  Gallon. 

268i 

u 

Inches 

a 

1  Dry  Gallon. 

2150xVo 

.  ii 

Inches 

a 

1  Bushel. 

128 

u 

Feet 

u 

1  Cord. 

Problem  I. 

6(31.  To  find  the  surface  of  a  right  prism. 
MiiUipIy  the  perimeter  of  the  base  by  the  altitude ,  and 
the  product  will  be  the  convex  surface    (Prop.  I.  Bk. 


282  ELEMENTS  OF  GEOMETRY. 

VIII.).      To  this  add  the  areas  of  the  tiuo  bases,  and  the 
result  ivilL  be  the  entire  surface. 

Examples. 

1.  Required  the  entire  surface  of  a 
pentangular  prism,  having  each  side  of 
its  base,  ABODE,  equal  to  2  feet,  and 
its  altitude,  AF,  equal  to  5  feet. 

2  X  5  =  10 ;  10  X  5  =  50  square  feet, 
[the  surface  required. 

2.  The  altitude  of  a  hexangular  prism  is  12  feet,  two  of 
its  faces  are  each  2  feet  wide,  three  are  each  2^  feet  wide, 
and  the  remaining  face  is  9  inches  wide ;  what  is  the 
convex  surface  of  the  prism  ? 

3.  Required  the  entire  surface  of  a  cube,  the  length  of 
each  edge  being  25  feet. 

4.  Required,  in  square  yards,  the  wall  surface  of  a  rec- 
tangular room,  whose  height  is  20  feet,  width  30  feet,  and 
length  50  feet.  Ans.  355t  sq.  yd. 

Problem  II. 

662.  To  find  the  solidity  of  a  prism. 
Multiply  the  area  of  its  base  by  its  altitude,  and  the 
product  will  be  its  solidity  (Prop.  XIII.  Bk.  YIII.). 

Examples. 

1.  Required  the  solidity  of  a  pentangular  prism,  having 
each  side  of  its  base  equal  to  2  feet,  and  its  altitude  equal 
to  5  feet. 

2^  X  1.72048  =  6.88192 ;  6.88192  X  5  =  34.40960  cubic 

[feet,  the  solidity  reqviired. 

2.  Required  the  solidity  of  a  triangular  prism,  whose 
length  is  10  feet,  and  the  three  sides  of  whose  base  are  3, 
4,  and  5  feet.  Ans.  60. 

3.  A  slab  of  marble  is  8  feet  long,  3  feet  wide,  and  6 
inches  tliick  ;  required  its  solidity. 


BOOK  XII.  283 

4.  There  is  a  cistern  in  the  form  of  a  cube,  whose  edge 
is  10  feet ;  what  is  its  capacity  in  liquid  gallons  ? 

Ans.  7480.519  gallons. 

5.  Required  the  solid  contents  of  a  quadrilateral  prism, 
the  length  being  19  feet,  the  sides  of  the  base  43,  54,  62, 
and  38,  and  the  diagonal  between  the  first  and  second 
sides,  70  inches.  Ans.  306.047  cu.  ft. 

6.  How  many  cords  in  a  range  of  wood  cut  4  feet  long, 
the  range  being  4  feet  6  inches  high  and  160  feet  long  ? 

Problem  III. 

663.  To  find  the  surface  of  a  right  pyramid. 
MuUiphj  the  perimeter  of  the  base  by  half  its  slant 

heig-ht,  and  the  product  ivill  be  the  convex  surface  (Prop. 
XV,  Bk.  VIII.).  To  this  add  the  area  of  the  base,  and 
the  result  ivill  be  the  entire  surface. 

664.  Scholium.  The  surface  of  an  oblique  pyramid  is 
found  by  taking  the  sum  of  the  areas  of  its  several  faces. 

S 
Examples. 

1.  Required  the  convex  surface  of 
a  pentangular  pyramid,  A  B  C  D  E  -  S, 
each  side  of  whose  base,  A  B  C  D  E,  is 
5  feet,  and  whose  slant  height,  S  M,  is 
20  feet. 
5  X  5  =  25 ;  25  X  -^^-  =  250  square 

[feet,  the  surface  required.  g         q 

2.  What  is  the  entire  surface  of  a  triangular  pyramid, 
of  which  the  slant  height  is  18  feet,  and  each  side  of  the 
base  42  inches  ?  Ans.  99.804  sq.  ft. 

3.  Required  the  convex  surface  of  a  triangular  pyramid, 
tlie  slant  height  being  20  feet,  and  each  side  of  the  base 
3  feet. 

4.  What  is  the  entire  surface  of  a  quadrangular  pyra- 
mid, the  sides  of  the  base  being  40  and  30  inches,  and  the 
slant  height  upon  the  greater  side  20.04,  and  upon  the  less 
side  20.07  feet  ?  Ans.  125.308  ft. 


284  ELEMENTS  OF  GEOMETRY. 

Problem  IV. 

665.  To   find   the   surface   of  a   frustum  of   a  right 

PYRAMID. 

Multiply  half  the  sum  of  the  perimeters  of  its  tvm  bases 
hy  its  slant  heig-ht,  and  the  product  will  be  the  convex 
surface  (Prop.  XVII.  Bk.  VIII.)  ;  to  this  add  the  areas 
of  the  two  bases f  and  the  result  will  be  the  entire  surface. 

Examples. 

1.  What  is  the  entire  surface  of  a  rectangular  frustum 
whose  slant  height  is  12  feet,  and  the  sides  of  whose  bases 
are  5  and  2  feet  ? 

5  X  4  =  20  ;  2X4  =  8;  20+  8  =  28  ;  2/  X  12=  168  ; 
52  +  22  =  29  ;  168  +  29  =  197  sq.  ft.,  area  required. 

2.  Required  the  convex  surface  of  a  regular  hexangular 
frustum,  whose  slant  height  is  16  feet,  and  the  sides  of 
whose  bases  are  2  feet  8  inches  and  3  feet  4  inches. 

3.  What  is  the  entire  surface  of  a  regular  pentangular 
frustum,  whose  slant  height  is  11  feet,  and  the  sides  of 
whose  bases  are  18  and  34  inches  ? 

Ans.  136.849  sq.  ft. 

Problem  V. 

666.  To  find  the  solidity  of  a  pyramid. 

Multiply  the  area  of  its  base  by  one  third  of  its  altitude 
(Prop.  XX.  Bk.  VIII.). 

Examples. 

1.  Required  the  solidity  of  a  pen- 
tangular pyramid,  A  B  C  D  E  -  S,  each 
side  of  whose  base,  ABODE,  is  5 
feet,  and  whose  altitude,  SO,  is  15 
feet. 

6^  X  1.7205  =  43.0125 ;  43.0125  X 
J^  =  215.0575  cu.  ft.,  the  solidity 
required. 


COOK  XII.  285 

2.  What  is  the  solidity  of  a  hexangular  pyramid,  the 
altitude  of  which  is  9  feet,  and  each  side  of  the  base  29 
inches  ? 

3.  What  is  the  solidity  of  a  square  pyramid,  each  side 
of  whose  base  is  30  feet,  and  whose  perpendicular  height 
is  25  feet  ?  Ans.  7500. 

4.  Required  the  solid  contents  of  a  triangular  pyramid, 
the  perpendicular  height  of  which  is  24  feet,  and  the  sides 
of  the  base  34,  42,  and  50  inches.      Ans.  39.2354  cu.  ft. 

Problem  YI. 

667.  To  find  the  solidity  of  a  frustum  op  a  pyramid. 

Add  together  the  areas  of  the  tivo  bases  and  a  mean 
proportional  beticeen  them,  and  multipli/  that  sum  by  one 
third  of  the  altitude  of  the  frustum  (Prop.  XXI.  Bk. 
YIIL). 

ExAMri-KS. 

1.  Required  the  solidity  of  the  frustum  of  a  quadran- 
gular pyramid,  the  sides  of  whose  bases  are  3  feet  and  2 
feet,  and  whose  altitude  is  15  feet. 

3X3  =  9;  2x2  =  4;  V"y><4=6  (Prop.  lY.  Bk.  11.)  ; 
(9-1-4  +  6)  X  V-  =  95  cu.  ft.,  solidity  required. 

2.  How  many  cubic  feet  in  a  stick  of  timber  in  the  form 
of  a  quadrangular  frustum,  the  sides  of  whose  bases  are 
15  inches  and  6  inches,  and  whose  altitude  is  20  feet  ? 

3.  Required  the  solid  contents  of  a  pentangular  frus- 
tum, whose  altitude  is  5  feet,  each  side  of  whose  lower 
base  is  18  inches,  and  each  side  of  whose  upper  base  is 
6  inches.  Ans.  9.319  cu.  ft. 

4.  Required  the  solidity  of  the  frustum  of  a  triangular 
pyramid,  the  altitude  of  which  is  14  feet,  the  sides  of  the 
lower  base  21,  15,  and  12,  and  those  of  the  upper  base  14, 
10,  and  8  feet.  Ans.  868.752  cu.  ft. 


286 


ELEMENTS   OF   GEOMETRY. 


The  Wedge. 

668.  A  Wedge  is  a  polyedron  bounded  by  a  rectangle, 
called  tlie  base  of  tlie  wedge  ;  by  two  trapezoids,  called  the 
sides,  which  meet  in  an  edge  parallel  to  the  base  ;  and  l)y 
two  triangles,  called  the  ends  of  the  wedge. 

Thus  ABCD-GH  is 
a  wedge,  of  which  ABCD 
is  the  rectangular  base ; 
ABHG,  DC  II G,  the  tra- 
pezoidal sides,  which  meet 
in  the  edge  GH ;  and  ADG, 
B  C  H,  the  triangular  ends. 

The  altitude  of  a  wedge  is  the  perpendicular  distance 
from  its  edge  to  the  plane  of  its  base  ;  as  G  P. 


Problem  YII. 

669.  To  find  the  solidity  of  a  wedge. 

Add  the  length  of  the  edg-e  to  twice  the  length  of  the 
base  ;  multiply  the  sum  by  one  sixth  of  the  product  of  the 
altitude  of  the  wedge  and  the  breadth  of  the  base. 

For,  let  L  equal  AB,  the 
lengtli  of  the  base  ;  /  equal 
GH,  the  length  of  the  edge ; 
b  equal  B  C,  the  breadth 
of  the  base  ;  and  h  equal 
PG,  the  height  of  the 
wedge.  Tlien  L  —  /  = 
A  B  —  G  H  =  A  M. 

Now,  if  the  length  of  tlie  base  and  the  edge  be  equal, 
the  polyedron  is  equal  to  half  a  parallclopipcdon  having 
the  same  base  and  altitude  (Prop.  VI.  Bk.  Vlll.),  and  its 
solidity  will  be  equal  to  ^  b  I  h  (Prop.  XHI.  Bk.  VIII.). 

If  the  length  of  the  base  is  greater  than  that  of  the 
edge,  let  a  section,  M  N  G,  be  made  parallel  to  B  C  11. 


BOOK  XII.  287 

This  section  will  divide  tlie  whole  wedge  into  the  quad- 
rangular pyramid  A  M  N  D  -  G,  and  the  triangular  prism 
BC^II-G. 

The  solidity  of  A  M  N  D  -  G  is  equal  to  ibhx  (L  —  /) 
(Prob.  V.)  ;  and  the  solidity  of  BCH-G  is  equal  to 
-J-  b  I  h  ;  hence  the  solidity  of  the  wliole  wedge  is  equal  to 

ibhl+ibh  X  (L  —  0  =  ibh^l+^bh'2L  — 
ibh  21=  ibh  X  (2L  +  0- 

But,  if  the  length  of  the  base  is  less  than  that  of  tlie 
edge,  the  solidity  of  the  wedge  will  be  equal  to  the  prism 
less  the  pyramid  ;  or  to 

^bhl  —  ^bhx  (l  —  L)  =  ^b  h  ^  I  —  ib  h  2  I  + 
hbh2L  =  ibhX  (2L  +  0. 

Examples. 

1.  Required  the  solidity  of  a  wedge,  the  edge  of  which 
is  10  inches,  the  sides  of  the  base  12  inches  and  6  inches, 
and  the  altitude  14  inches. 

10  4-  (12  X  2)  =  34 ;  34  X  ^^^  =  476  cu.  in.,  the 

[solidity  required. 

2.  "What  is  the  solidity  of  a  wedge,  of  which  the  edge  is 
24  inches,  the  sides  of  the  base  36  inches  and  9  inches, 
and  the  altitude  22  inches  ? 

3.  How  many  solid  feet  in  a  wedge,  of  which  the  sides 
of  tlie  base  are  35  inches  and  15  inches,  the  length  of  the 
edge  55  inches,  and  the  altitude  17/o  inches  ? 

Ans.  3  cu.  ft.  17of  cu.  in. 

Rectangular  Prismoid. 

670.  A  RECTANGULAR  PRISMOID  is  a  polyedron  bounded 
by  two  rectangles,  called  the  bases  of  the  prismoid,  and  by 
four  trapezoids  called  the  lateral  faces  of  the  prismoid. 

The  altitude  of  a  prismoid  is  the  perpendicular  distance 
between  its  bases. 


288  elements  op  geometry. 

Problem  YIII. 

671.  To  find  the  solidity  of  a  rectangular  prtsmotd. 

Add  the  area  of  the  tvm  bases  to  four  times  the  area  of 
a  parallel  section  at  equal  distances  from  the  bases  ;  mul- 
tiply the  sum  by  one  sixth  of  the  altitude. 

Let  L  and  B  be  the  length  and  breadth       J^:^ — i k 

of  the  lower  base,  /  and  b  the  length  and       /  \ l-A 

/III 
breadth  of  the  upper  base,  M  and  m  tlie      f''^"'\  [\ 

length  and  breadth  of  the  parallel  section     / "  t 

equidistant   from   the   bases,  and  h  the       \^/  \| 

altitude  of  the  prismoid.  

If  a  plane  be  passed  through  tlie  opposite  edges  L  and  /, 
the  prismoid  will  be  divided  into  two  wedges,  having  for 
bases  the  bases  of  the  prismoid,  and  for  edges  L  and  /. 

The  solidity  of  these  wedges,  which  compose  the  pris- 
moid, is  (Prob.  Vll.), 

^BAX  (2L  +  /)  +  iZ;Ax  (2Z+L)  =  iA(2BL  + 
But  M  being  equally  distant  from  L  and  /,  2  M  =  L  +  ^, 
and  2m  =  B  +  ^  (Prop.  YII.  Cor.,  Bk.  lY.)  ;  conse- 
quently, 

4Mm=(L  +  /)X(B  +  ^^)  =  BL  +  B/  +  ^^L  +  Z^/. 

Substituting  4M7yi  for  its  base,  in  the  preceding  equation, 
we  have,  as  the  expression  of  the  solidity  of  a  prismoid, 

^h  (BL-f  ^Z  +  4Mm). 

672.  Scholium.  This  demonstration  applies  to  prismoids 
of  other  forms.  For,  whatever  be  the  form  of  the  two 
bases,  there  may  be  inscribed  in  each  such  a  number  of 
small  rectangles  that  the  sum  of  them  in  each  base  shall 
differ  less  from  that  base  than  any  assignable  quantity  ;  so 
that  tlie  sum  of  the  rectangular  prismoids  that  may  be 


BOOK  XII.  289 

constructed  on  these  rectangles  will  differ  from  the  given 
prismoid  by  less  than  any  assignable  quantity. 

Examples. 

1.  Kcquired  the  solidity  of  a  prismoid,  tlie  larger  base 
of  which  is  80  inches  by  27  inclies,  the  smaller  base  24 
inches  by  18  inches,  and  the  altitude  48  inches. 

QA  _j_  24  27  -4-  18 

SO  X  27  ==  810  ;  21  X  18  =  432 ;  ~\~  X  — ^  X  4 
=  2430  ;  (810  +  432  +  2430)  X  ¥  =  29,376  cu.  in. 
=  17  cu.  ft.,  the  solidity  required. 

2.  What  is  the  solidity  of  a  stick  of  timber,  whose  larger 
end  is  24  inches  by  20  inches,  the  smaller  end  16  inches 
by  12  inches,  and  the  length  18  feet  ? 

3.  What  is  the  solidity  of  a  block,  whose  ends  are  re- 
spectively 30  by  27  inches  and  24  by  18  inches,  and  whose 
length  is  36  inches  ? 

4.  What  is  the  capacity  in  gallons  of  a  cistern  47:|^  inches 
deep,  whose  inside  dimensions  are,  at  the  top  81^  and  55 
inches,  and  at  the  bottom  41  and  29^  inches  ? 

Ans.  546.929  gall. 

Problem  IX. 

673.  To  find  the  surface  of  a  regular  polyedron. 

Multiphj  the  area  of  one  of  the  faces  by  the  number  of 
faces ;  or  multiply  the  square  of  one  of  the  edges  of  the 
polyedron  by  the  surface  of  a  similar  polyedron  whose 
edges  are  1. 

For,  since  the  faces  of  a  regular  polyedron  are  all  equal, 
it  is  evident  that  the  area  of  one  face  multiplied  by  the 
number  of  faces  will  give  the  area  of  the  whole  surface. 
Also,  since  the  surfaces  of  regular  polyedrons  of  the  same 
name  are  bounded  by  the  same  number  of  similar  poly- 
gons (Prop.  I.  Bk.  VI.),  their  surfaces  are  to  each  other 
as  the  squares  of  the  edges  of  the  polyedrons  (Prop.  I. 
Cor.,  Bk.  VI.). 

25 


290 


ELEMENTS   OF   GEOMETRY. 


674.     Table  of  Surfaces  and  Solidities  of  Polyedrons 
WHOSE  Edge  is  1. 


NAMES. 

NO.  OF  FACES. 

SURFACES. 

SOI.ID1TIES. 

Tetraedron, 

Hexaedron, 

Octaedron, 

Dodecaedron, 

Icosaedron, 

4 
6 

8 
12 

20 

1.7320508 
6.0000000 
3.4641016 
20.6457288 
8.6602540 

0.1178511 
1.0000000 
0.4714045 
7.603118a 
2.1816050 

The  surfaces  in  the  table  are  obtained  by  multiplying 
the  area  of  one  of  the  faces  of  the  polyedron,  as  given  in 
Art.  632,  by  the  number  of  faces. 

Examples. 

1.  What  is  the  surface  of  an  octaedron  whose  edge  is  16 
inches  ? 

16'  X  3.4641016  --  886.81  sq.  in.,  the  area  required. 

2.  Required  the  surface  of  an  icosaedron  whose  edge  is 
20  inches. 

3.  Required  the  surface  of  a  dodecaedron  whose  edge  is 
12  feet.  Ans.  2972.985  sq.  ft. 

Problem  X. 

6T5.  To  find  the  solidity  of  a  regular  polyedron. 

Multiply  the  surface  by  one  third  of  the  perpendicular 
distance  from  the  centre  to  one  of  the  faces  ^  or  multiply 
the  cube  of  one  of  the  edg-es  by  the  solidity  of  a  similar 
polyedron  ivhose  edge  is  1. 

For  any  regular  polyedron  may  be  divided  into  as  many 
equal  pyramids  as  it  has  faces,  the  common  vertex  of  the 
pyramids  being  the  centre  of  the  polyedron  ;  lience,  the 
solidity  of  the  polyedron  must  equal  the  product  of  the 
areas  of  all  its  faces  by  one  third  the  perpendicular  dis- 
tance from  the  centre  to  each  face  of  the  polyedron. 


BOOK    XII. 


201 


Also,  since  similar  pyramids  arc  to  each  other  as  the 
cubes  of  their  homologous  edges  (Prop.  XXII.  Bk.  VIIL), 
two  polyedrons  containing  the  same  number  of  similar 
pyramids  are  to  each  other  as  the  cubes  of  their  edges  ; 
hence,  the  solidity  of  a  polyedron  whose  edge  is  1  (Art. 
673),  may  be  used  to  measure  other  similar  polyedrons. 

Examples. 

1.  Required  the  solidity  of  an  octaedron  whose  edge  is 
16  inches. 

IG'  X  0.4714045  =  1930.8728  cu.in.,  solidity  required. 

2.  What  is  the  solidity  of  a  tetraedron  whose  edge  is 
2  feet  ? 

3.  Required  the  solidity  of  au  icosaedron  whose  edge 
is  15  inches.  Aiis.  7303. 220G  cii.  in. 


pROnLEM    XI. 

676.  To  find  tlie  surface  of  a  cylinder. 

MuUipIfj  the  circumference  of  its  base  by  its  altitude^ 
and  the  product  will  be  the  convex  surface  (Prop.  I.  Bk. 
X.).  To  this  add  the  areas  of  its  two  bases,  and  the  re- 
sult will  be  the  entire  surface. 


Examples. 

1.  What  is  the  entire  surface  of  a  cylin- 
der, tlie  altitude  of  wliicli,  AB,  is  10  feet, 
and  the  circumference  of  the  base  20  feet  ? 

10  X  20  =  200  ;  20^  x  0.07958  X  2  = 
63.264;  200  +  63.264  =  263.264  sq.  ft., 
the  surface  required. 

2.  Required  the  convex  surface  of  a  cylinder  whose  alti- 
tude is  16  feet,  and  the  circumference  of  whose  base  is  21 
feet. 

3.  Wliat  is  the  entire  surface  of  a  cylinder  whose  alti- 
tude is  10  inches,  and  whose  circumference  is  4  feet  ? 


292  ELEMENTS  OP  GEOMETRY. 

4.  How  many  times  must  a  cylinder  5  feet  3  inches 
long,  and  21  inches  in  diameter,  revolve,  to  roll  an  acre  ? 

Ans.  1509.18  times. 

Problem  XII. 

677.  To  find  the  solidity  of  a  cylinder. 

Multiply  the  area  of  the  base  by  the  altitude^  and  the 
product  will  be  the  solidity  (Prop.  II.  Bk.  X.). 

Examples. 

1.  What  is  the  solidity  of  a  cylinder,  whose  altitude  is 
10  feet,  and  the  circumference  of  whose  base  is  20  feet  ? 

20^  X  0.07958  X  10  ==  318.32  cu.  ft.,  solidity  required. 

2.  Hequired  the  solidity  of  a  cylindrical  log,  whose  length 
is  9  feet,  and  the  circumference  of  wliose  base  is  6  feet. 

Ans.  25.7831  cu.  ft. 

3.  The  Winclicster  bushel  is  a  liollow  cylinder  18j- 
inches  in  diameter,  and  8  inches  deep ;  what  is  its  ca- 
pacity in  cubic  inches  ? 

Problem  XIII.       ^ 

678.  To  find  the  surface  of  a  cone. 

Multiply  the  circumference  of  the  base  by  half  the  slant 
heig-ht  (Prob.  III.  Bk.  X.),  and  the  product  will  be  the 
convex  surface.  To  this  add  the  area  of  the  base,  and  the 
result  will  be  the  entire  surface. 

Examples. 

1.  What  is  the  convex  surface  of  a  cone,  whose  slant 
height  is  28  feet,  and  the  circumference  of  whose  base  is 
40  feet  ? 

40  X  ^/  =  560  sq.  ft.,  the  surface  required. 

2.  Bequired  tlie  entire  surface  of  a  cone,  whose  slant 
height  is  14  feet,  and  the  circumference  of  whose  base  is 
92  inches. 


BOOK  XII.  293 

3.  Wliat  is  the  surface  of  a  cone,  whose  slant  height  is 
9  feet,  and  the  diameter  of  whose  base  is  36  inches  ? 

4.  How  many  yards  of  canvas  are  required  for  the  cov- 
ering of  a  conical  tent,  the  slant  height  of  which  is  30  feet, 
and  the  circumference  of  the  base  900  feet  ? 

Ans.  1500  sq.  yd. 

Problem  XIY. 

679.  To  find  the  surface  of  a  frustum  of  a  cone. 

MuUiphj  half  the  sum  of  the  circumferences  of  its  tivo 
bases  hy  its  slant  hei^-ht,  and  the  product  ivill  be  the  con- 
vex surface  (Prop.  IV.  Bk.  X.).  To  this  add  the  area 
of  its  bases,  and  the  result  will  he  the  entire  surface. 

G80.  Scholium.  The  convex  surface  of  a  frustum  of  a 
cone  may  also  be  found  by  multiplying  the  slant  height  by 
the  circumference  of  a  section  at  equal  distances  between 
the  two  bases  (Prop.  lY.  Cor.,  Bli.  X.). 

Examples. 

1.  Required  the  convex  surface  of  a  frustum  of  a  cone, 
whose  slant  height  is  20  feet,  and  the  circumferences  of 
wliose  bases  are  30  feet  and  40  feet. 

— "-^ —  X  20  =  700  sq.  ft.,  the  surface  required. 

2.  Required  tlie  surface  of  a  frustum  of  a  cone,  the  di- 
ameters of  the  bases  being  43  inches  and  23  inches,  and 
the  slant  height  9  feet. 

3.  What  is  the  convex  surface  of  a  frustum  of  a  cone, 
of  which  a  section  equidistant  from  its  two  bases  is  24  feet 
in  circumference,  the  slant  height  of  X\\q  frustum  being 
19  feet  ? 

4.  From  a  cone  the  circumference  of  wliose  base  is  10 
feet,  and  whose  slant  heiglit  is  CO  feet,  a  cone  has  been 
cut  off,  whose  slant  height  is  8  feet.  What  is  the  convex 
surface  of  the  frustum  ?  Ans.  139^  sq.  ft. 


294  ELEMENTS  OP  GEOMETRY. 

Problem  XV. 

681.  To  find  the  solidity  of  a  cone. 

MuUiply  the  area  of  its  base  by  one  third  of  its  altitude, 
and  the  product  will  be  the  solidity  (Prop.  Y.  Bk.  X.). 

Examples. 

1.  What  is  tlie  solidity  of  a  cone  whose  altitude  is  42 
feet,  and  the  diameter  of  wliose  base  is  10  feet  ? 

102  X  0.7854  X  ¥-  =  1099.56  cu.  ft.,  solidity  required. 

2.  Required  tlie  solidity  of  a  cone  wliose  altitude  is  63 
feet,  and  the  radius  of  whose  base  is  12  feet  6  inches. 

8.  Plow  many  cubic  feet  in  a  conical  stick  of  timber, 
whose  length  is  18  feet,  the  diameter  at  the  larger  end 
being  42  inches  ?  Ans.  57.7269  cu.  ft. 

Problem  XVI. 

682.  To  fmd  the  solidity  of  the  frustum  of  a  cone. 
Add  together  the  areas  of  the  two  bases  and  a  mean 

proportional  bettoeen  them,  and  midliply  that  sum  by  one 
third  of  the  altitude  of  the  frustum  ;  and  the  result  ivill  be 
the  solidity  required  (Prop.  VI.  Ek.  X.). 

Examples. 

1.  What  is  the  solidity  of  a  frustum 
of  a  cone,  C  D  E  F,  whose  altitnde, 
A  B,  is  21  feet,  and  the  area  of  whose 
bases,  F  E,  CD,  are  80  square  feet 
and  300  square  feet  ? 

(80  +  300  +  V80  X  300)   X  -V  = 
3732.96  cu.  ft.,  solidity  required. 

2.  Required  the  solidity  of  a  frustum  of  a  cone,  the 
diameters  of  the  bases  being  38  and  27  inches,  and  the 
altitude  11  feet. 

3.  If  a  cask,  which  is  two  equal  frustums  of  cones  joined 
together  at  the  larger  bases,  have  its  bung  diameter  28 


LOOK    XII. 


295 


inches,  the  head  diameter  20  inches,  and  length  40  inches, 
how  many  gallons  of  wine  will  it  hold  ?         Ans.  79.06. 

Problem  XVII. 

683,  To  find  the  surface  of  a  sphere. 

Multiply  the  diameter  by  the  circumference  of  a  great 
circle  of  the  sphere  (Prop.  VIII.  Bk.  X.)  ;  or  multiply 
the  area  of  one  great  circle  of  the  sphere  by  4  (Proj). 
VIII.  Cor  1,  Bk.  X.)  ;  or  multiply  3.1416  by  the  square 
of  the  diameter  (Prop.  VIII.  Cor.  4,  Bk.  X.). 

D 

Examples. 

1.  What  is  the  surface  of 
a  sphere,  whose  diameter,  ED, 
is  40  feet,  and  whose  circum- 
ference, A  E  B  D,  is  125.664? 

125.664  X  40  =  5026.56  sq. 
[ft.,  the  surface  required. 


2.  Required  the  surface  of  a  sphere  whose  diameter  is 
30  inches. 

3.  What  is  the  surface  of  a  globe  whose  diameter  is  7 
feet  and  circumference  21.99  feet  ?  Ans.  153.93. 

4.  How  many  square  miles  of  surface  has  the  earth,  its 
diameter  being  7912  miles  ? 


Problem  XVIII. 
684.  To  find  the  surface  of  a  zone  or  segment  op  a 

SPHERE. 

Multiply  the  altitude  of  the  zone  or  segment  by  the  cir- 
cumference  of  a  great  circle  of  the  sphere  (Prop.  VIII. 
Cor.  2,  Bk.  X.)  ;  or  mtdtiply  the  product  of  the  diameter 
and  altitude   by  3.1416  (Prop.  VIII.  Cor.  6,  Bk.  X.). 


296  ELEMENTS   OF   GEOMETRY. 

EXAMPLKS. 

1.  What  is  the  surface  of  a  segment  of  a  sphere,  the 
altitude  of  the  segment  being  10  feet,  and  the  diameter  of 
the  sphere  50  feet  ? 

50  X  10  X  3.1416  =  1570.80  sq.  ft.,  surface  required. 

2.  The  altitude  of  a  segment  of  a  spliere  is  88  inches, 
and  the  circumference  of  the  sphere  is  25  feet ;  Avhat  is 
the  surface  of  the  segment  ? 

3.  Required  the  surface  of  a  zone  or  segment,  the  diam- 
eter of  the  sphere  being  72  feet,  and  the  altitude  of  the 
zone  24  feet.  Ans.  5428.6848  sq.  ft. 

4.  If  the  earth  be  regarded  as  a  perfect  sphere  whose 
axis  is  7912  miles,  and  tlie  part  of  the  axis  corresponding 
to  each  of  the  frigid  zones  is  327.192848,  to  each  of  the 
temperate  zones  2053.468612,  and  to  the  torrid  zone 
8150.67708  miles  ;  what  is  the  surface  of  each  zone  ? 
Ans.  Each  frigid  zone  8132797.39568  ;  each  temperate  zone 

51041592.99898;  torrid  zone  78314115.07768  miles. 

Problem  XIX. 

685.  To  find  the  solidity  of  a  sphere. 

Multiply  the  surface  of  the  sphere  by  one  third  of  its 
radius  (Prop.  IX.  Bk.  X.)  ;  or  midtiply  the  cube  of  the  di- 
ameter of  the  sphere  by  0.5236  (Prop.  IX.  Cor.  5,  Bk.  X.). 

Examples. 

1.  What  is  the  solidity  of  a  sphere  whose  diameter  is 
40  inches  ? 

40^  X  0.5236  =  33510.4  cu.  in.,  the  solidity  required. 

2.  Required  the  solidity  of  a  globe  whose  circumference 
is  60  inches. 

3.  What  is  the  solidity  of  the  moon  in  cubic  miles,  sup- 
posing it  a  perfect  sphere  with  a  diameter  of  2160  miles  ? 

4.  Required  the  solidity  of  the  eartli,  supposing  it  to  Co 
a  perfect  sphere,  whose  diameter  is  7912  miles. 

Ans.  259332805349.80493  cu.  miles. 


BOOK   XII. 


207 


Problem  XX. 

686.  To  fiiid  the  surface  of  a  spherical  polygon. 

From  the  sum  of  all  the  angles  subtract  the  product  of 
two  right  angles  by  the  number  of  sides  less  tivo  ;  divide 
the  remaimler  by  90°,  and  multiply  the  quotient  by  one 
eighth  of  the  surface  of  the  sphere ;  and  the  result  ivill  be 
the  surface  of  the  spherical  polygon  (Prop.  XX.  Bk.  IX.). 

Examples. 

1.  Required  the  surface  of  a  spherical  polygon  having 
five  sides,  described  on  a  sphere  whose  diameter  is  100 
feet,  tlie  sum  of  the  angles  being  720  degrees. 

2  X  90°  X  (5  —  2)  =  540° ;  (720°  —  540°)  -^-  90°  =  2 ; 
100^  X  3.1416  =  31416  ;  2  X  ^-^F^  =  7854  sq.  ft.,  the 
surface  required. 

2.  What  is  the  surface  of  a  triangle  on  a  sphere  whose 
diameter  is  20  feet,  the  angles  being  150°,  90°,  and  54°  ? 


Problem  XXI. 
687.  To  find  the  solidity  of  a  spherical  pyramid  or 

SECTOR. 

Multiply  the  area  of  the  polygon  or  zone  which  forms 
the  base  of  the  pyramid  or  sector  by  one  third  of  the  radius 
(Prop.  IX.  Cor.  1,  Bk.  X.)  ;  or  multiply  the  altitude  of 
the  base  by  the  square  of  the  radius,  and  that  product  by 
2.0944  (Prop.  IX.  Cor.  7,  Bk.  X.). 

Examples. 
1.  Required  the  solidity  of  a 
spherical  sector,  A  C  B  E,  the  al- 
titude, ED,  of  the  zone  forming 
its  base  being  5  feet,  and  the 
radius,  C  B,  of  the  sphere  being 
12  feet. 

5  X  24  X  3.1416  =  376.992  ; 
376.992  X  -V-  =  1507.968  cu. 
ft.,  the  solidity  required. 


298  ELEMENTS   OF   GEOMETRY. 

2.  What  is  the  solidity  of  a  spherical  pyramid,  the  area 
of  its  base  being  3G4  square  feet,  and  the  diameter  of  the 
sphere  60  feet  ? 

3.  Required  the  solidity  of  a  spherical  sector,  whose 
base  is  a  zone  16  inches  in  altitude,  in  a  sphere  3  feet  in 
diameter. 

4.  What  is  the  solidity  of  a  spherical  sector,  whose  base 
is  a  zone  6  feet  in  altitude,  iu  a  sphere  18  feet  in  diam- 
eter ?  Ans.  1017.88  cu.  ft. 

Problem  XXII. 

688.  To  find  the  solidity  of  a  spherical  segment. 
When  the  segment  is  less  than  a  hemisphere^  from  the 

solidity  of  the  spherical  sector  luhose  base  is  the  zone  of 
the  seg-ment,  take  the  soliditij  of  the  cone  ivhose  vertex  is 
the  centre  of  the  sphere^  and  ivhose  base  is  the  circular 
base  of  the  segment;  but  ivhen  the  segment  is  greater 
than  a  hemisphere,  take  the  sum  of  these  solidities  (Prop. 
IX.  Sch.,  Bk.  X.). 

689.  Scholium,  If  the  segment  has  two  plane  bases,  its 
solidity  may  be  found  by  taking-  the  difference  of  the  tiro 
segments  which  lie  on  the  same  side  of  its  two  bases 
(Prop.  IX.  Sch.,  Bk.  X.). 

Examples. 

1.  What  is  the    solidity  of  a 
segment,  ABE,  whose  altitude, 
E  D,  is  5  feet,  cut  from  a  sphere 
whose  radius,  C  E,  is  20  feet  ? 
The  altitude  of  the  cone  A  B  C  is 

equal  to  CE  —  ED,  or  20  —  5, 

which  is  equal  to  15  feet ;  and 

the  radius  of  its  base  is  equal  to 

VC A2"-i:CD^  or  V20^  — 15S 

which  is  equal  to  13.23 ;   consequently   the  diameter 

A  B  is  equal  to  26.46  feet ;  5  X  20^  X  2.0944  =  4188.8 


BOOK  XII.  299 

cubic  feet,  the  solidity  of  the  sector  A  C  D  E  (Prob. 
XXI.)  ;  20.40^  X  0.7854  X  -V'-  =  294.6,99  cubic  feet, 
the  solidity  of  the  cone  A  B  -  C  (Prob.  XV.)  ;  4188.8 
—  2946.99  ==  1241.81  cubic  feet,  the  solidity  of  the 
segment  ABE  required. 

2.  Required  tlie  solidity  of  a  segment,  whose  altitude  is 
57  inches,  the  diameter  of  the  sphere  being  153  inches. 

3.  What  is  the  solidity  of  a  spherical  segment,  whose 
altitude  is  13  feet,  and  tlie  diameter  of  the  sphere  S3  feet 
6  inches  ? 

4.  Required  the  solidity  of  the  segments  of  the  earth 
which  are  bounded  severally  by  its  five  zones,  the  earth's 
diameter  being  7912  miles,  and  the  part  of  the  diameter 
corresponding  to  each  of  the  frigid  zones  being  327.19, 
to  each  temperate  zone  2053.47,  and  to  the  torrid  zone 
3150.68. 

Ans.  Each  frigid  zone  1293793463.32,  each  temperate  zone 
55013912318.45,  and  the  torrid  zone  146717393786.26 
cubic  miles. 

The  Spheroid. 

690.  A  SPHEROID  is  a  solid  which  may  be  described  by 
the  revolution  of  an  ellipse  about  one  of  its  axes,  which 
remains  immovable. 

An  oblate  spheroid  is  one  described  by  the  revolution  of 
the  ellipse  about  its  minor  or  conjugate  axis. 

A  prolate  spheroid  is  one  described  by  the  revolution  of 
the  ellipse  about  its  major  or  transverse  axis. 

Problem  XXIII. 

691.  To  find  the  solidity  of  a  spheroid. 

Multiply  the  square  of  the  axis  of  revolution  by  the  fixed 
axis,  and  that  product  by  0.5 236. 

A  full  demonstration  of  this,  which  is  based  upon  the 
principle  that  a  spheroid  is  two  thirds  of  its  circumscribing 


800  ELEMENTS   OF  GEOMETRY. 

cylinder,  would  require  a  knowledge  of  Conic  Sections,  or 
of  the  DitTerential  and  Integral  Calculi,  with  neither  of 
which  is  the  learner  here  supposed  to  be  acquainted. 

The  relation,  however,  of  the  spheroid  to  its  circumscrib- 
ing cylinder,  is  that  which  the  sphere  sustains  to  its  cir- 
cumscribing cylinder  (Prop.  X.  Bk.  X.). 

Now  the  area  of  the  base  of  the  cylinder  is  fovmd  by 
multiplying  the  square  of  the  axis  of  revolution  by  0.7854, 
and  the  solidity  of  the  cylinder  by  multiplying  that  pro- 
duct by  the  fixed  axis  (Prop.  II.  Bk.  X.).  But  the  solid- 
ity of  the  spheroid  is  only  two  thirds  of  that  of  the  cylin- 
der ;  hence,  to  obtain  the  solidity  of  the  former,  instead  of 
multiplying  by  0.7854,  we  must  use  a  factor  only  two 
thirds  as  large,  which  will  be  0.5236. 

Examples. 

1.  What  is  the  solidity  of  the  ob- 
late spheroid  ACBD,  whose  fixed 
axis,  C  D,  is  30  inches,  and  the  axis 
of  revokition,  A  B,  40  inches. 
402  X  30  X  0.5236  =  25132.8  cubic 

inches,  the  solidity  required. 

2.  Required  the  solidity  of  a  prolate  spheroid,  whose 
fixed  axis  is  50  feet,  and  the  axis  of  revolution  36  feet. 

3.  What  is  the  solidity  of  a  prolate,  and  also  of  an  oblate 
spheroid,  the  axes  of  each  being  25  and  15  inches  ? 

Ans.  Prolate,  2945.25  cu.  in. ;  oblate,  4908.75  cu.  in. 

4.  What  is  the  solidity  of  a  prolate,  and  also  of  an  ob- 
late spheroid,  the  axes  of  each  being  3  feet  6  inches  and 
2  feet  10  inches  ? 

5.  Required  the  solidity  of  the  earth,  its  figure  being 
that  of  an  oblate  spheroid  whose  axes  are  7925.3  and 
7898.9  miles.  Ans.  259774584886.834  cubic  miles. 


BOOK    XIII. 

MISCELLANEOUS    GEOMETRICAL    EXERCISES. 

1.  If  the  opposite  angles  formed  by  four  lines  meeting 
at  a  point  are  equal,  these  lines  form  but  two  straight 
lines. 

2.  If  the  equal  sides  of  an  isosceles  triangle  are  pro- 
duced, the  two  exterior  angles  formed  with  the  base  will 
be  equal. 

3.  The  sum  of  any  two  sides  of  a  triangle  is  greater  than 
the  third  side. 

4.  If  from  any  point  within  a  triangle  two  straight  lines 
are  drawn  to  the  extremities  of  either  side,  they  will  in- 
clude a  greater  angle  than  that  contained  by  the  other 
two  sides. 

5.  If  two  quadrilaterals  have  the  four  sides  of  the  one 
equal  to  the  four  sides  of  the  otlier,  each  to  each,  and  the 
angle  included  by  any  two  sides  of  tlie  one  equal  to  the 
angle  contained  by  the  corresponding  sides  of  the  other, 
the  quadrilaterals  are  themselves  equal. 

6.  The  sum  of  the  diagonals  of  a  trapezium  is  less  than 
the  sum  of  any  four  lines  which  can  be  drawn  to  the  four 
angles  from  any  point  within  the  figure,  except  from  the 
intersection  of  the  diagonals. 

7.  Lines  joining  the  corresponding  extremities  of  two 
equal  and  parallel  straight  lines,  are  themselves  equal  and 
parallel,  and  the  figure  formed  is  a  parallelogram. 

8.  If,  in  the  sides  of  a  square,  at  equal  distances  from 
the  four  angles,  points  be  taken,  one  in  each  side,  the 
straight  lines  joining  these  points  will  form  a  square. 

26 


302  ELEMENTS   OP   GEOMETRY. 

9.  If  one  angle  of  a  parallelogram  is  a  right  angle,  all 
its  angles  are  riglit  angles. 

10.  Any  straight  line  drawn  throngli  the  middle  point 
of  a  diagonal  of  a  parallelogram  to  meet  the  sides,  is  bi- 
sected in  that  point,  and  likewise  bisects  the  parallelogram. 

11.  If  four  magnitudes  are  proportionals,  the  first  and 
second  may  be  multiplied  or  divided  by  the  same  magni- 
tude, and  also  the  third  and  fourth  by  the  same  magni- 
tude, and  the  resulting  magnitudes  will  be  proportionals. 

12.  If  four  magnitudes  are  proportionals,  the  first  and 
third  may  be  multiplied  or  divided  by  the  same  magni- 
tude, and  also  the  second  and  fourth  by  the  same  magni- 
tude, and  the  resulting  magnitudes  will  be  proportionals. 

13.  If  there  be  two  sets  of  proportional  magnitudes,  the 
quotients  of  the  corresponding  terms  will  be  proportionals. 

14.  If  any  two  points  be  taken  in  the  circumference  of 
a  circle,  the  straight  line  joining  them  will  lie  wholly 
within  the  circle. 

15.  The  diameter  is  the  longest  straight  line  that  can 
be  inscribed  in  a  circle. 

16.  If  two  straight  lines  intercept  equal  arcs  of  a  circle, 
and  do  not  cut  each  other  within  the  circle,  the  lines  will 
be  parallel. 

17.  If  a  straight  line  be  drawn  to  touch  a  circle,  arfd  be 
parallel  to  a  chord,  the  point  of  contact  Avill  be  the  middle 
point  of  the  arc  cut  off  by  that  chord. 

18.  If  two  circles  cut  each  other,  and  from  either  point 
of  intersection  diameters  be  drawn,  the  extremities  of  these 
diameters  and  the  other  point  of  intersection  will  be  in  the 
same  straight  line. 

19.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be 
the  diameter  of  a  circle,  the  circumference  of  the  circle 
will  bisect  the  base  of  the  triangle. 

20.  If  the  opposite  angles  of  a  quadrilateral  be  together 
equal  to  two  right  angles,  a  circle  may  be  circumscribed 
about  the  quadrilateral. 


BOOK  XIII.  303 

21.  Parallelograms  wliicli  have  two  sides  and  the  in- 
cluded angle  equal  in  each,  are  themselves  equal. 

22.  Equivalent  triangles  upon  the  same  base,  and  upon 
the  same  side  of  it,  are  between  the  same  parallels. 

23.  If  the  middle  points  of  the  sides  of  a  trapezoid, 
which  are  not  parallel,  be  joined  by  a  straight  line,  that 
line  will  be  parallel  to  each  of  the  two  jjarallel  sides,  and 
be  equal  to  half  their  sum. 

24.  If,  in  opposite  sides  of  a  parallelogram,  at  equal 
distances  from  opposite  angles,  points  be  taken,  one  in 
each  side,  the  straight  line  joining  these  points  will  bisect 
the  parallelogram. 

25.  The  perimeter  of  an  isosceles  triangle  is  greater 
than  the  perimeter  of  a  rectangle,  which  is  of  the  same 
altitude  with,  and  equivalent  to,  the  given  triangle. 

26.  If  the  sides  of  the  square  described  upon  the  hypoth- 
enuse  of  a  right-angled  triangle  be  produced  to  meet  the 
sides  (produced  if  necessary)  of  the  squares  described 
upon  the  other  two  sides  of  the  triangle,  the  triangles 
thus  formed  will  be  similar  to  the  given  triangle,  and 
two  of  them  will  be  equal  to  it. 

27.  A  square  circumscribed  about  a  given  circle  is 
double  a  square  inscribed  in  the  same  circle. 

28.  If  the  sum  of  the  squares  of  the  four  sides  of  a 
quadrilateral  be  equivalent  to  the  sum  of  the  squares  of 
the  two  diagonals,  the  figure  is  a  parallelogram. 

29.  Straight  lines  drawn  from  the  vertices  of  a  triangle, 
so  as  to  bisect  the  opposite  sides,  bisect  also  the  triangle. 

30.  The  straight  lines  which  bisect  the  three  angles  of 
a  triangle  meet  in  the  same  point. 

31.  The  area  of  a  triangle  is  ecfual  to  its  perimeter  mul- 
tiplied by  half  the  radius  of  the  inscribed  circle. 

32.  If  the  points  of  bisection  of  the  sides  of  a  given  tri- 
angle be  joined,  the  triangle  so  formed  will  be  one  fourth 
of  the  given  triangle. 

33.  To  describe  a  square  upon  a  given  straight  line. 


304  ELEMENTS  OF  GEOMETRY. 

34.  To  find  in  a  given  straight  line  a  point  equally  dis- 
tant from  two  given  points. 

35.  To  construct  a  triangle,  the  base,  one  of  the  angles 
at  the  base,  and  the  sum  of  the  other  two  sides  being 
given. 

36.  To  trisect  a  right  angle. 

37.  To  divide  a  triangle  into  two  parts  by  a  line  drawn 
parallel  to  a  side,  so  that  these  parts  shall  be  to  each  other 
as  two  given  straight  lines. 

38.  To  divide  a  triangle  into  two  parts  by  a  line  drawn 
perpendicular  to  tlie  base,  so  that  these  parts  shall  be  to 
each  other  as  two  given  lines. 

39.  To  divide  a  triangle  into  two  parts  by  a  line  drawn 
from  a  given  point  in  one  of  the  sides,  so  that  the  parts 
shall  be  to  each  other  as  two  given  lines. 

40.  To  divide  a  triangle  into  a  square  number  of  equal 
triangles,  similar  to  eacli  other  and  to  the  original  triangle. 

41.  To  trisect  a  given  straight  line. 

42.  To  inscribe  a  square  in  a  given  right-angled 
isosceles  triangle. 

43.  To  inscribe  a  square  in  a  given  quadrant. 

44.  To  describe  a  circle  that  shall  pass  through  a  given 
point,  have  a  given  radius,  and  touch  a  given  straight  line. 

45.  To  describe  a  circle,  the  centre  of  wliich  shall  be  in 
the  perpendicular  of  a  given  right-angled  triangle,  and  the 
circumference  of  which  shall  pass  through  the  right  angle 
and  touch  the  hypotlienuse. 

46.  To  describe  three  circles  of  equal  diameters  which 
shall  touch  each  other,  and  to  describe  another  circle 
wliicli  shall  touch  the  three  circles. 

47.  If,  on  the  diameter  of  a  semicircle,  two  equal  circles 
be  described,  and  in  the  curvilinear  space  included  by  the 
three  circumferences  a  circle  be  inscribed,  its  diameter 
will  be  to  that  of  the  equal  circles  in  the  ratio  of  two  to 
three. 

48.  If  two  points  be  taken  in  tlie  diameter  of  a  circle, 


BOOK  XIII.  305 

equidistant  from  the  centre,  the  sum  of  tlie  squares  of  two 
lines  drawn  from  these  points  to  any  point  in  the  circum- 
ference will  always  be  the  same. 

49.  Given  the  vertical  angle,  and  the  radii  of  the  in- 
scribed and  circumscribed  circles,  to  construct  the  triangle. 

50.  If  a  diagonal  cuts  off  three,  five,  or  any  odd  number 
of  sides  from  a  regular  polygon,  the  diagonal  is  parallel  to 
one  of  the  sides. 

51.  The  area  of  a  regular  hexagon  inscribed  in  a  circle 
is  double  that  of  an  equilateral  triangle  inscribed  in  the 
same  circle. 

52.  The  side  of  a  square  circumscribed  about  a  circle 
is  equal  to  the  diagonal  of  a  square  inscribed  in  the  same 
circle. 

53.  To  describe  a  circle  equal  to  half  a  given  circle. 

54.  A  regular  duodecagon  is  equivalent  to  three  fourths 
of  the  square  constructed  on  the  diameter  of  its  circum- 
ocribed  circle  ;  or  is  equal  to  the  square  constructed  on 
the  side  of  the  equilateral  triangle  inscribed  in  the  same 
circle. 

55.  If  semicircles  be  described  on  the  sides  of  a  right- 
angled  triangle  as  diameters,  the  one  described  on  the 
hypothenuse  will  be  equal  to  the  sum  of  the  other  two. 

56.  If  on  the  sides  of  a  triangle  inscribed  in  a  semi- 
circle, semicircles  be  described,  the  two  crescents  thus 
formed  will  together  equal  the  area  of  the  triangle. 

57.  If  the  diameter  of  a  semicircle  be  divided  into  any 
number  of  parts,  and  on  them  semicircles  be  described, 
their  circumferences  will  together  be  equal  to  the  circum- 
ference of  the  given  semicircle. 

58.  To  divide  a  circle  into  any  number  of  parts,  which 
shall  all  be  equal  in  area  and  equal  in  perimeter,  and  not 
have  the  parts  in  the  form  of  sectors. 

59.  To  draw  a  straight  line  perpendicular  to  a  plane, 
from  a  given  point  above  the  plane. 

60.  Two  straight  lines  not  in  the  same  plane  being 

20* 


306  ELEMENTS   OF   GEOMETRY. 

given  ill  position,  to  draw  a  straight  line  which  shall  be 
perpendicular  to  them  both. 

61.  The  solidity  of  a  triangular  prism  is  equal  to  the 
product  of  the  area  of  either  of  its  rectangular  sides  as  a 
base  multiplied  by  half  its  altitude  on  that  base. 

62.  All  prisms  of  equal  bases  and  altitudes  are  equal  in 
solidity,  whatever  be  tiie  figure  of  tiieir  bases. 

63.  The  convex  surface  of  a  regular  pyramid  exceeds 
the  area  of  its  base  in  the  ratio  that  the  slant  height  of  the 
pyramid  exceeds  the  radius  of  the  circle  inscribed  in  its 
base. 

64.  If  from  any  point  in  the  circumference  of  the  base 
of  a  cylinder,  a  straight  line  be  drawn  perpendicular  to 
the  plane  of  the  base,  it  will  be  wholly  in  the  surface  of 
the  cylinder. 

65.  A  cylinder  and  a  parallelopipedon  of  equal  bases 
and  altitudes  are  equivalent  to  each  other. 

6Q.  If  two  solids  have  the  same  height,  and  if  their  sec- 
tions made  at  equal  altitudes,  by  planes  parallel  to  the 
bases,  have  always  the  same  ratio  which  the  bases  have 
to  one  another,  the  solids  have  to  one  another  the  same 
ratio  which  their  bases  have. 

67.  The  side  of  the  largest  cube  that  can  be  inscribed 
in  a  sphere,  is  equal  to  the  square  root  of  one  tliird  of  the 
square  of  the  diameter  of  the  sphere. 

68.  To  cut  off  just  a  square  yard  from  a  plank  14  feet 
3  inches  long,  and  of  a  uniform  width,  at  what  distance 
from  the  edge  must  a  line  be  struck  ?  Ans.  7^^  in. 

69.  How  much  carpeting  a  yard  wide  will  be  required 
to  cover  the  floor  of  an  octagonal  hall,  whose  sides  arc  10 
feet  each  ? 

70.  The  perambulator,  or  surveying-wheel,  is  so  con- 
structed as  to  turn  just  twice  in  the  length  of  a  rod  ;  what 
is  its  diameter  ?  Ans.  2.626  ft. 

71.  What  is  the  excess  of  a  floor  50  feet  long  by  00 
broad,  above  two  others,  each  of  half  its  dimensions  ? 


BOOK  XIII.  307 

72.  The  four  sides  of  a  trapezium  are  13,  13.4,  24,  and 
18  feet,  and  the  first  two  contain  a  right  angle.  Required 
tlie  area.  Ans.  253.38  sq.  ft. 

73.  If  an  equilateral  triangle,  whose  area  is  equal  to 
10,000  square  feet,  be  surrounded  with  a  walk  of  uniform 
width,  and  equal  to  the  area  of  the  mscribed  circle,  what 
is  the  width  of  the  walk  ?  Ans.  11.701  ft. 

74.  A  right-angled  triangle  has  its  base  16  rods,  and  its 
perpendicular  12  rods,  and  a  triangle  is  cut  off  from  it  by  a 
line  parallel  to  its  base,  of  which  the  area  is  24  rods.  Re- 
quired the  sides  of  tliat  triangle.     Ans.  8,  6,  and  10  rods. 

75.  There  is  a  circular  pond  whose  area  is  5028i  square 
feet,  in  the  middle  of  which  stood  a  pole  100  feet  high ; 
now,  the  pole  having  been  broken  off,  it  was  observed  that 
the  top  portion  resting  on  the  stump  just  reached  the  brink 
of  the  pond.  What  is  the  height  of  the  piece  left  stand- 
ing ?  Ans.  41.9968  ft. 

76.  The  area  of  a  square  inscribed  in  a  circle  is  400 
square  feet ;  required  the  diagonal  of  a  square  circum- 
scribed about  the  same  circle. 

77.  The  four  sides  of  a  field,  whose  diagonals  are  equal, 
are  known  to  be  25,  35,  31,  and  19  rods,  in  a  successive 
order  ;  what  is  the  area  of  the  field  ? 

Ans.  4A.  IR.  38^  p. 

78.  The  wheels  of  a  chaise,  each  4  feet  high,  in  turning 
within  a  ring,  moved  so  that  the  outer  wheel  made  two 
turns  while  the  inner  made  one,  and  their  distance  from 
one  another  was  5  feet ;  what  were  the  circumferences  of 
the  tracks  described  by  them  ? 

Ans.  Outer,  62.8318  ft.;  inner,  31.4159  ft. 

79.  The  girt  of  a  vessel  round  the  outside  of  the  hoop 
is  22  inches,  and  the  hoop  is  1  inch  thick ;  required  the 
true  girt  of  the  vessel. 

80.  If  one  of  the  Egyptian  pyramids  is  490  feet  high, 
having  each  slant  side  an  equilateral  triangle  and  the  base 
a  square,  what  is  the  area  of  the  base  ? 

Ans.  11  A.  3  rd.  2234  ft. 


308  ELEMENTS  OF  GEOMETRY. 

81.  An  ellipse  is  surrounded  by  a  wall  14  inclies  thick ; 
its  axes  are  840  links  and  612  links  ;  required  the  quan- 
tity of  ground  enclosed,  and  the  quantity  occupied  by  the 
wall. 

Ans.  4  A.  6  rd.  enclosed,  and  1760.49  sq.  ft.,  area  oc- 
cupied by  the  wall. 

82.  There  is  a  meadow  of  1  acre  in  the  form  of  a  square ; 
what  must  be  the  length  of  the  rope  by  which  a  horse,  tied 
equidistant  from  eacli  angle,  can  be  permitted  to  graze 
over  the  entire  meadow  ? 

83.  A  gentleman  has  a  rectangular  garden,  whose  length 
is  100  feet  and  breadth  80  feet ;  what  must  be  the  uni- 
form width  of  a  walk  half-way  round  the  same,  to  take 
up  just  half  the  garden  ?  Ans.  25.9688  ft. 

84.  Two  trees,  100  feet  asunder,  are  placed,  the  one  at 
the  distance  of  100  feet,  and  the  other  50  feet  from  a  wall ; 
what  is  the  distance  that  a  person  must  pass  over  in  run- 
ning from  one  tree  to  touch  the  wall,  and  then  to  the  other 
tree,  the  lines  of  distance  making  equal  angles  with  the 
wall  ?  Ans.  173.205  ft. 

85.  There  is  a  rectangular  park  400  feet  long  and  300  feet 
l)road,  all  round  which,  and  close  by  the  wall,  is  a  border 
10  feet  broad ;  close  by  the  border  there  is  a  walk,  and 
also  two  others,  crossing  each  other  and  the  park  at  right 
angles,  m  the  middle  of  the  garden.  The  walks  are  all 
of  one  breadth,  and  their  area  takes  up  one  tenth  of  the 
whole  park  ;  required  the  breadth  of  the  walks. 

Ans.  6.2375  ft. 

86.  A  farmer  borrowed  a  cubical  pile  of  Avood,  which 
measured  6  feet  every  way,  and  repaid  it  by  two  cubical 
piles,  of  which  the  sides  were  3  feet  each ;  what  part  of 
the  quantity  borrowed  lias  he  returned  ? 

87.  A  board  is  10  feet  long,  8  inches  in  breadth  at  the 
greater  end,  and  6  inches  at  the  less ;  how  much  must 
be  cut  off  from  tlie  less  end  to  make  a  square  foot  ? 

Ans.  23.2493  in. 

88.  A  piece  of  timber  is  10  feet  long,  each  side  of  the 


BOOK  XIII.  309 

greater  base  9  inches,  and  each  side  of  tlie  less  6  inches  ; 
how  much  must  be  cut  off  from  the  less  end  to  contain  a 
solid  foot?  Ans.  3.39214ft. 

89.  What  must  be  the  inside  dimensions  of  a  cubical 
box  to  hold  200  balls,  each  2^  inches  in  diameter  ? 

90.  Near  my  house  I  intend  making  a  hexagonal  or  six- 
sided  seat  around  a  tree,  for  which  I  have  procured  a  pine 
plank  16|^  feet  long  and  11  inches  broad  ;  what  must  be 
the  inner  and  outer  lengths  of  each  side  of  the  seat,  that 
there  may  be  the  least  loss  in  cutting  up  tlie  plank  ? 

Ans.  26.64915  in.  inner,  and  39.35085  in.  outer  length. 

91.  Required  the  capacity  of  a  tub  in  the  form  of  a 
frustum  of  a  cone,  of  which  the  greatest  diameter  is  48 
inches,  the  inside  length  of  the  staves  30  inches,  and  the 
diagonal  between  the  farthest  extremities  of  the  diameters 
50  inches.  Ans.  165.34  gals. 

92.  The  front  of  a  house  is  of  such  a  height,  that,  if  the 
foot  of  a  ladder  of  a  certain  length  be  placed  at  the  dis- 
tance of  12  feet  from  it,  the  top  of  the  ladder  will  just 
reach  to  the  top  of  the  house ;  but  if  the  foot  of  the  ladder 
be  placed  20  feet  from  the  front,  its  top  will  fall  4  feet  be- 
low the  top  of  the  house.  Required  the  height  of  the  house, 
and  the  length  of  the  ladder. 

Ans.  34  feet,  the  height  of  the  building ;  36.0555  feet, 
the  length  of  the  ladder. 

93.  A  sugar-loaf  in  form  of  a  cone  is  20  inches  high  ;  it 
is  required  to  divide  it  equally  among  three  persons,  by  sec- 
tions parallel  to  the  base  ;  what  is  the  height  of  each  part  ? 

Ans.  Upper  13.8672,  next  3.6044,  lowest  2.5284  in. 

94.  Within  a  rectangular  court,  whose  length  is  four 
chains,  and  breadth  three  chains,  there  is  a  piece  of  water 
in  the  form  of  a  trapezium,  whose  opposite  angles  are  in  a 
direct  line  with  those  of  the  court,  and  the  respective  dis- 
tances of  tlie  angles  of  the  one  from  those  of  the  other  are 
20,  25,  40,  and  45  yards,  in  a  successive  order  ;  required 
the  area  of  the  water.  Ans.  960  sq.  yd. 


310  ELEMENTS  OF  GEOMETRY. 

95.  What  will  the  diameter  of  a  sphere  be,  when  its 
solidity  and  the  area  of  its  surface  are  expressed  by  the 
same  numbers  ?  Ans.  6. 

96.  There  is  a  circular  fortification,  which  occupies  a 
quarter  of  an  acre  of  ground,  surrounded  by  a  ditch  coin- 
ciding with  the  circumference,  24  feet  wide  at  bottom,  2G 
at  top,  and  12  deep  ;  how  much  water  will  fill  the  ditch, 
if  it  slope  equally  on  both  sides  ?     Ans.  135483.25  cu.  ft. 

97.  A  father,  dying,  left  a  square  field  containing  CO 
acres  to  be  divided  among  his  five  sons,  in  such  a  manner 
that  the  oldest  son  may  have  8  aci-es,  the  second  7,  the 
third  6,  the  fourth  5,  and  the  fifth  4  acres.  Now,  tlic 
division  fences  are  to  be  so  made  that  the  oldest  son's 
share  shall  be  a  narrow  piece  of  equal  breadth  all  around 
the  field,  leaving  the  remaining  four  shares  in  the  form  of 
a  square  ;  and  in  like  manner  for  each  of  the  other  shares, 
leaving  always  the  remainders  in  form  of  squares,  one 
within  another,  till  tlie  share  of  the  youngest  be  the  inner- 
most square  of  all,  equal  to  4  acres.  Ilcquired  a  side  of 
each  of  the  enclosures. 

Ans.  17.3205, 14.8324, 12.2474,  9.4868,  and  6.3246  chains. 

98.  Required  the  dimensions  of  a  cone,  its  solidity  be- 
ing 282  inches,  and  its  slant  height  being  to  its  base  diam- 
eter as  5  to  4. 

Ans.  9.796  in.  the  base  diameter ;  12.246  in.  the  slant 
height;  and  11.223  in.  the  altitude. 

99.  A  gentleman  has  a  piece  of  ground  in  form  of  a 
square,  the  difference  between  whose  side  and  diagonal  is 
10  rods.  He  would  convert  two  thirds  of  the  area  into  a 
garden  of  an  octagonal  form,  but  would  have  a  fish-pond 
at  the  centre  of  tlie  garden,  in  the  form  of  an  equilateral 
triangle,  whose  area  must  equal  five  square  rods.  Re- 
quired the  length  of  each  side  of  the  garden,  and  of  each 
side  of  the  pond. 

Ans.  8.9707  rods,  each  side  of  the  garden,  and  3.398 
rods,  each  side  of  the  pond. 


BOOK    XIV. 


APPLICATIONS   OF   ALGEBRA  TO   GEOMETRY. 

692.  When  it  is  proposed  to  solve  a  geometrical  prob- 
lem by  aid  of  Algebra,  draw  a  figure  wbich  shall  represent 
the  several  parts  or  conditions  of  the  problem,  both  known 
and  required. 

Represent  the  known  parts  by  the  first  letters  of  the 
alphabet,  and  the  required  parts  by  the  last  letters. 

Then,  observing  the  geometrical  relations  that  the  parts 
of  the  figure  have  to  each  other,  make  as  many  indepen- 
dent equations  as  there  are  unknown  quantities  intro- 
duced, and  the  solution  of  these  equations  will  determine 
the  unknown  quantities  or  required  parts. 

To  form  these  equations,  howeyer,  no  definite  rules  can 
be  given  ;  but  the  best  aids  may  be  derived  from  ex[)eri- 
ence,  and  a  thorough  knowledge  of  geometrical  principles. 

It  should  be  the  aim  of  the  learner  to  effect  the  simplest 
solution  possible  of  each  problem. 

Problem  I. 

693.  In  a  right-aiigled  triangle,  having'  given  the  hy- 
pothenuse,  and  the  sum  of  the  other  two  sides,  to  deter- 
mine these  sides. 

C 

Let  A  B  C  be  the  triangle,  right-an- 
gled at  B.  Put  A  C  :=j  a,  the  sum  A  B 
+  B  C  =  5,  AB  =.  .T,  and  B  C  =  2/- 


312 


ELEMENTS   OF   GEOMETRY. 


Then,  x  -\-  1/  =  s, 

and  (Prop.  XI.  Bk.  lY.), 

x'  +  7/  =  a\ 
From  the  first  equation,         x  =^  s  —  p. 

Substitute  in  second  equation  this  vahie  of  .r, 

Or,  2p^  —  2sij=^  a'  —  s% 

Or,  ,f^s7j  =  ia'—is\ 

By  completing  the  square, 

2/'  —  si/  +  is'=^a^  —  is', 

Extracting  sq.  root,      1/  —  ^  5  =  ±  V  ^  a-  —  ^  s^^ 

Or,  7/  =  ^s  ±  V^a'—is\ 

If  A  C  =  5,  and  the  sum  AB  +  BC  =  7,  2/  =  4or3, 
and  X  =  3  or  4. 

Problem  II. 

694.  Having  given  the  base  and  perpendicular  of  a  tri- 
angle^ to  find  the  side  of  an  inscribed  square. 

Let   ABC    be   the    triangle,  C 

and  H  E  F  G  the  inscribed 
square.  Put  A  B  =  6,  C  D  =  a, 
and  GForOH  =  DI  =  x-; 
then  will  CI  =  CD  —  DI  = 
a  —  X. 

Since    the    triangles    ABC,     A 
Gr  F  C  are  similar, 

AB:CD::GF:CI, 
or  b  :  a  :  :  X  :  a  —  x. 

Hence,  a  b  —  b  x  =  a  .x, 

a  b 

or,  *  =  «  +  &• 


DE    B 


BOOK   XTV.  313 

that  is,  the  side  of  the  inscribed  square  is  equal  to  the  pro- 
duct of  the  base  by  the  altitude ,  divided  by  their  sum. 

Problem  III. 

695.  Having'  given  the  lengths  of  two  straight  lines 
drawn  from  the  acute  angles  of  a  right-angled  trian- 
gle to  the  7niddle  of  the  opposite  sides,  to  determine  those 
sides. 

Let  A  B  C  be  tho  given  triangle,  A 

and  A  D,  B  E  the  given  lines. 

Put  A  D  =  «,  B E  =  Z;,  CD  or  ^ 
CB  =  .^,  and  CE  or  ^  CA  =  y; 
then,  since  C  D^  +  C  A=^  =  A  D\  and  /     / ^\P' 

CE^+CB^  =  BES 
we  have        x"^  -)-  4  y^  =  a^. 


and  y^  +  4  x^  =  b^,  B  D  C 

By  subtracting  the  second  equation  from  four  timQs  the 
first, 

lby^=4:a^  —  b% 


jfi  —  h^ 


J2 


by  subtracting   the   first   equation   from  four  times  the 
second, 

15  x^  =  4.b^  —  a\ 


"I 


or,  x=    \^^-_^', 

15 

which  values  of  x  and  y  are  half  the  base  and  perpendic- 
ulars of  the  triangle. 

Problem  IV. 

696.  In  an  equilateral  triangle,  having  given  the  lengths 
of  the  three  perpendiculars  drawn  from  a  point  vnthin  to 
the  three  sides,  to  determine  these  sides. 

27 


314  ELEMENTS   OF   GEOMETRY. 

Let  A  B  C  be  the  equilateral  trian- 
gle, and  D  E,  D  F,  D  G  the  given  per- 
pendiculars from  the  point  D.  Draw 
D  A,  D  B,  D  C  to  the  vertices  of  the 
three  angles,  and  let  fall  the  perpen- 
dicular, OH,  on  the  base,  AB. 

Put  DE  =  6t,  DF  =  6,  DG  =  c,     ^  ^^        ^ 

and  A  H  or  B  H,  half  the  side  of  the  equilateral  triangle, 
=  X.     Then  A  C  or  B  C  =  2  .r ,  and  C  H  =  ^ KQ^^^KW 
=  \f\x^  —  o;^  =  V  8  x,"^  =  a;  V  3.     Now,  since  the  area 
of  a  triangle  is  equal  to  the  product  of  half  its  base  by  its 
altitude  (Prop.  VI.  Bk.  IV.), 
The  triangle  ACB  =  J^  A B X  CH  =  x X x^/l^^x'^l, 
ABD  =  ^ABxBG  =  .^Xc        =cx, 
BCD  =  ^BCxr)E=a;X^        =ax. 
ACD  =  ^ACxDF=a;X^>        =hx.        . 
But  the  three  triangles  ABD,  BCD,ACD  are  together 
equaf  to  the  triangle  A  C  B. 

Hence,  x^  ^  '^  =  ax  -\-  b  x  -\-  c  x  =  x  (a -\- h  -\-  c) , 

or,  .^•  V  3  =  «  +  &  +  c  ; 

a  4-&  -\-c 
or,  X  =       ^     . 

Hence  each  side,  or  2  a;  =  -^ . 

697.  Cor,  Since  the  perpendicular,  CH,  is  equal  to 
X  V  3,  it  is  equal  io  a  -\-  b  -\-  c ;  that  is,  the  ivhole  per- 
pendicular of  an  equilateral  triang-le  is  equal  to  the  sum 
of  all  the  perpendiculars  let  fall  from  any  point  in  the  tri- 
angle to  each  of  its  sides. 

Problem  V. 

698.  To  determine  the  radii  of  three  equal  circles  de- 
scribed within  and  tangent  to  a  given  circle,  and  also  tan- 
iX'ent  to  each  other. 


BOOK  XIV. 


315 


Let  A  F  be  the  radius  of  the 
given  circle,  and  B  E  the  radius 
of  one  of  the  equal  circles  de- 
scribed within  it.  Put  A¥  =  a, 
and  BE  =  x;  then  each  side  of 
the  equilateral  triangle,  BCD, 
formed  by  joining  the  centres  of 
the  required  circles,  will  be  rep- 
resented by  2x,  and  its  altitude, 
C  E,  by  V 47^3^^%  or  x  Vs. 

The  triangles  B  C  E,  ABE  are  similar,  since  the  angles 
B  C  E  and  ABE  are  equal,  each  being  half  as  great  as 
one  of  the  angles  of  the  equilateral  triangle,  and  the  angle 
B  E  C  is  common. 


Hence, 

CE  :BE::BC;  AB, 

or 

xV~S:x::2x:  AB, 

and 

AB_^!. 

V3 

But 

AB  +  BF  =  AF; 

hence, 

2x 

or 

2x  +  x^S  =  aV^, 

or 

(2  -f  VS)  x  =  aV^, 

Hence, 

^- ;/;.- 2.1. -»x  0.4641. 

Problem  YI. 

699.  In  a  right-angled  triangle,  having  given  the  base, 
and  the  sum  of  the  perpendicular  and  hypothenuse,  to 
find  these  two  sides. 


Problem  YH. 

YOO.  In  a  rectangle,  having  given   the  diagonal   and 
perimeter,  to  find  the  sides. 


316  elements  of  geometry. 

Problem  YIII. 

701.  In  a  riglit-aiigled  triangle,  having  given  the  base, 
and  the  difference  between  the  hypothenuse  and  perpen- 
dicular, to  find  both  these  two  sides. 

Problem  IX. 

702.  Having  given  the  area  of  a  rectangle  inscribed  in 
a  given  triangle,  to  determine  the  sides  of  the  rectangle. 

Problem  X. 
.    703.  In  a  triangle,  having  given  the  ratio  of  the  two 
sides,  together  with  both»  the  segments  of  the  base,  made 
by  a  perpendicular  from  the  vertical  angle,  to  determine 
the  sides  of  the  triangle. 

Problem  XI. 

704.  In  a  triangle,  having  given  the  base,  the  sum  of 
the  other  two  sides,  and  the  length  of  a  line  drawn  from 
the  vertical  angle  to  the  middle  of  the  base,  to  find  the 
sides  of  the  triangle. 

Problem  XII. 

705.  In  a  triangle,  having  given  the  two  sides  about 
the  vertical  angle  together  with  the  line  bisecting  that 
angle,  and  terminating  in  the  base,  to  find  the  base. 

Problem  XIII. 

706.  To  determine  a  right-angled  triangle,  having  given 
the  perimeter  and  the  radius  of  its  inscribed  circle. 

Problem  XIY. 

707.  To  determine  a  triangle,  having  given  the  base, 
the  perpendicular,  and  the  ratio  of  the  two  sides. 

Problem  XY. 

708.  To  determine  a  right-angled  triangle,  having  given 
the  hypothenuse,  and  the  side  of  the  inscribed  square. 


BOOK   XIV.  317 

Problem  XYI. 

709.  In  a  right-angled  triangle,  having  given  the  perim- 
eter, or  sum  of  all  the  sides,  and  the  perpendicular  let 
fall  from  the  right  angle  on  the  hypothenuse,  to  determine 
the  triangle,  that  is,  its  sides. 

Problem  XYII. 

710.  To  determine  a  right-angled  triangle,  having  given 
the  hypothenuse,  and  the  difference  of  two  lines  drawn 
from  the  two  acute  angles  to  the  centre  of  the  inscribed 
circle. 

'     Problem  XYIII. 

711.  To  determine  a  triangle,  having  given  the  base,  the 
perpendicular,  and  the  difference  of  the  two  other  sides. 

Problem  XIX. 

712.  To  determine  a  triangle,  having  given  the  lengths 
of  three  lines  drawn  from  the  three  angles  to  the  middle 
of  the  opposite  sides. 

Problem  XX. 

713.  In  a  triangle,  having  given  all  the  three  sides,  to 
find  the  radius  of  the  inscribed  circle. 

Problem  XXI. 

714.  To  determine  a  right-angled  triangle,  having  given 
the  side  of  the  inscribed  square,  and  the  radius  of  the 
inscribed  circle. 

Problem  XXII. 

715.  To  determine  a  triangle,  having  given  the  base, 
tlie  perpendicular,  and  the  rectangle  of  the  two  other 
sideg. 

27* 


L 


318  ELEMENTS  OP  GEOMETRY. 

Problem  XXIII. 

716.  To  determine  a  right-angled  triangle,  having  given 
the  hypothenuse,  and  the  radius  of  the  inscribed  circle. 

Problem  XXIY. 

717.  To  determine  a  right-angled  triangle,  having  given 
the  hypothenuse  and  the  difference  between  a  side  and  the 
radius  of  the  inscribed  circle. 

Problem  XXY. 

718.  To  determine  a  triangle,  having  given  the  base, 
the  line  bisecting  the  vertical  angle,  and  the  diameter  of 
the  circumscribing  circle. 

Problem  XXVI. 

719.  There  are  two  stone  pillars  in  a  garden,  whose 
perpendicular  heights  are  20  and  30  feet,  and  the  distance 
between  them  60  feet.  A  ladder  is  to  be  placed  at  a  cer- 
tain point  in  the  line  of  distance,  of  such  a  length,  that 
it  may  just  reach  the  top  of  both  the  pillars.  What  is  the 
length  of  the  ladder,  and  how  far  from  each  pillar  must 
it  be  placed  ? 

Ans.  39.5899  feet,  length  of  the  ladder ;  34J  feet,  dis- 
tance of  the  foot  of  the  ladder  from  the  bottom  of  the 
lower  pillar ;  and  25f  feet,  distance  of  the  foot  of  the 
ladder  from  the  bottom  of  the  higher  pillar. 

Problem  XXVII. 

720.  There  is  a  cistern,  the  sum  of  the  length  and 
breadth  of  which  is  84  inches,  the  diagonal  of  the  top  60 
inches,  and  the  ratio  of  the  breadth  to  the  depth  as  25  to 
7.  What  are  its  dimensions,  provided  it  has  the  form  of 
a  rectangular  parallelopipedon  ? 

Ans.  Length  48  inches  ;  width  36  inches  ;  depth  10.08 
inches. 


BOOK  XIV.  319 

Problem  XXYIII. 

721.  The  three  distances  from  an  oak,  growing  in  an 
open  plain,  to  the  three  visihle  corners  of  a  square  field,' 
lyin^at  some  distance,  are  known  to  be  78,  59.161,  and 
78  poles,  in  successive  order.  What  are  the  dimensions 
of  the  field,  and  its  area  ? 

Ans.  Side  of  the  square  21  rd. ;  area  3  A.  2  R.  16  rd. 

Problem  XXIX. 

722.  There  is  a  house  of  three  equal  stories  in  height. 
Now  a  ladder  being  raised  against  it,  at  20  feet  distance 
from  the  foot  of  the  building,  reaches  the  top ;  whilst 
another  ladder,  12  feet  shorter,  raised  from  the  same 
point,  reaches  only  to  the  top  of  the  second  story.  What 
is  the  height  of  the  building  ?  Ans.  41.696  ft. 

Problem  XXX. 

723.  The  solidity  of  a  cone  is  2513.28  cubic  inches,  and 
the  slant  side  of  a  frustum  of  it,  whose  solidity  is  2474.01, 
is  19.5  inches.     Required  the  dimensions  of  the  cone. 

Ans.  Altitude  24  inches ;  base  diameter  20  inches. 

Problem  XXXI. 

724.  Within  a  rectangular  garden  containing  just  an 
acre  of  ground,  I  have  a  circular  fountain,  whose  circum- 
ference is  40,  28,  52,  and  60  yards  distant  from  the  four 
angles  of  the  garden.  From  these  dimensions,  the  length 
and  breadth  of  the  garden,  and  likewise  the  diameter  of 
the  fountain,  are  required. 

Ans.  Length  94.996  yds. ;  width  50.949  yds. ;  diameter 
of  the  fountain  20  yds. 

Problem  XXXII. 

725.  There  is  a  vessel  in  the  form  of  a  frustum  of  a 
cone,  standing  on  its  lesser  base,  whose  solidity  is  8.67 
feet,  the  depth  21  inches,  its  greater  base  diameter  to  that 


320  ELEMENTS  OF  GEOMETRY. 

of  the  lesser  as  7  to  5,  into  whicli  a  globe  had  accidentally 
been  put,  whose  solidity  was  2i  times  the  measure  of  its 
surface.  Required  the  diameters  of  the  vessel  and  of  the 
globe,  and  how  many  gallons  of  water  would  be  requisite 
just  to  cover  the  latter  within  the  former. 

A.ns.  35  and  25  inches,  top  and  bottom  diameters  of  tlie 

frustum ;  15  inches,  diameter  of  the  globe  ;  and  34.2 

gallons,  the  water  required. 

Problem  XXXIII. 

726.  Three  trees,  A,  B,  C,  whose  respective  heights  are 
114,  110,  and  98  feet,  are  standing  on  a  horizontal  plane, 
and  the  distance  from  A  to  B  is  112,  from  B  to  C  is 
104,  and  from  A  to  C  is  120  feet.  What  is  the  distance 
from  the  top  of  each  tree  to  a  point  in  the  plane  which 
shall  be  equally  distant  from  eacli  ?         Ans.  126.634  ft. 

Problem  XXXIV. 

727.  A  person  possessed  a  rectangular  meadow,  the 
fences  of  which  had  been  destroyed,  and  the  only  mark 
left  was  an  oak-tree  in  the  east  corner ;  he  however  recol- 
lected the  following  particulars  of  the  dimensions.  It 
had  once  been  resolved  to  divide  the  meadow  into  two 
parts  by  a  hedge  running  diagonally  ;  and  he  recollected 
that  a  segment  of  the  diagonal  intercepted  by  a  perpen- 
dicular from  one  of  the  corners  was  16  chains,  and  the 
same  perpendicular,  produced  2  chains,  met  the  other  side 
of  the  meadow.  Now  the  owner  has  bequeathed  it  to 
four  grandchildren,  whose  shares  are  to  be  bounded  by 
the  diagonal  and  perpendicular  produced.  What  is  the 
area  of  the  meadow,  and  what  are  the  several  shares  ? 

Ans.  Area  of  the  whole  meadow,  16  acres ;  shares,  1  B,. 
24  rd.;  1  A.  2  R.  16  rd. ;  6  A.  IB.  24  rd. ;  7  A. 
2  B.  16  rd. 

THE    END 


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